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Unformatted text preview: Graduate Texts in Mathematics 158 Editorial Board S. Axler K.A. Ribet Graduate Texts in Mathematics
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 TAKEUTI]ZARING. Introduction to Axiomatic Set Theory. 2nd ed. OXTOBY.Measure and Category. 2nd ed. SCHAEFER. Topological Vector Spaces. 2nd ed. HILTON/STAMMBACH. Course in A Homological Algebra. 2nd ed. MAC LANE. Categories for the Working Mathematician. 2nd ed. HUGHES/PIPER.Projective Planes. J.P. SERRE. A Course in Arithmetic. TAKEUTI/ZARING. Axiomatic Set Theory. HUMPHREYS.Introduction to Lie Algebras and Representation Theory. COHEN.A Course in Simple Homotopy Theory. CONWAY.Functions of One Complex Variable I. 2nd ed. BEALS.Advanced Mathematical Analysis. ANDERSON/FULLER.Rings and Categories of Modules. 2nd ed. GOLUBITSKY/GUILLEMIN. Stable Mappings and Their Singularities. BERBERIAN.Lectures in Functional Analysis and Operator Theory. WINTER.The Structure of Fields. ROSENBLATr.Random Processes. 2nd ed. HALMOS.Measure Theory. HALMOS.A Hilbert Space Problem Book. 2nd ed. HUSEMOLLER.Fibre Bundles. 3rd ed. HUMPHREYS.Linear Algebraic Groups. BARNES/MACK.An Algebraic Introduction to Mathematical Logic. GREUB. Linear Algebra. 4th ed. HOLMES.Geometric Functional Analysis and Its Applications. HEWITT/STROMBERG. Real and Abstract Analysis. MANES.Algebraic Theories. KELLEY.General Topology. ZARISKI]SAMUEL. Commutative Algebra. Vol. I. ZARISKI/SAMUEL. Commutative Algebra. Vol. II. JACOBSON.Lectures in Abstract Algebra I. Basic Concepts. JACOBSON.Lectures in Abstract Algebra II. Linear Algebra. JACOBSON.Lectures in Abstract Algebra III. Theory of Fields and Galois Theory. HIRSCH. Differential Topology. 34 35 36 37 38 39 40 41 SPITZER.Principles of Random Walk. 2nd ed. ALEXANDER/WERMER.Several Complex Variables and Banach Algebras. 3rd ed. KELLEY/NAMIOKA al. Linear et Topological Spaces. MONK. Mathematical Logic. GRAUERT/FRITZSCHE.Several Complex Variables. ARVESON.An Invitation to C*Algebras. KEMENY/SNELL/KNAPP.Denumerable Markov Chains. 2nd ed. APOSTOL.Modular Functions and Dirichlet Series in Number Theory. 2nd ed. J.P. SERRE. Linear Representations of Finite Groups. GILLMAN/JERISON. Rings of Continuous Functions. KENDIG.Elementary Algebraic Geometry. LOEVE.Probability Theory I. 4th ed. LOEVE.Probability Theory II. 4th ed. MOISE.Geometric Topology in Dimensions 2 and 3. SACHS/WU.General Relativity for Mathematicians. GRUENBERG/WEIR.Linear Geometry. 2nd ed. EDWARDS.Fermat's Last Theorem. KLINGENBERG. Course in Differential A Geometry. HARTSHORNE.Algebraic Geometry. MANIN.A Course in Mathematical Logic. GRAVER/WATKINS. Combinatorics with Emphasis on the Theory of Graphs. BROWN/PEARCY.Introduction to Operator Theory I: Elements of Functional Analysis. MASSEY.Algebraic Topology: An Introduction. CROWELL/FOX.Introduction to Knot Theory. KOBLITZ. padic Numbers, padic Analysis, and ZetaFunctions. 2nd ed. LANG.Cyclotomic Fields. ARNOLD.Mathematical Methods in Classical Mechanics. 2nd ed. WHITEHEAD.Elements of Homotopy Theory. KARGAPOLOV/MERIZJAKOV. Fundamentals of the Theory of Groups. BOLLOBAS.Graph Theory. 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 33 (continued after index) Steven Roman Field Theory
Second Edition With 18 Illustrations Springer To Donna Preface This book presents the basic theory of fields, starting more or less from the beginning. It is suitable for a graduate course in field theory, or independent study. The reader is expected to have taken an undergraduate course in abstract algebra, not so much for the material it contains but in order to gain a certain level of mathematical maturity. The book begins with a preliminary chapter (Chapter 0), which is designed to be quickly scanned or skipped and used as a reference if needed. The remainder of the book is divided into three parts. Part 1, entitled Field Extensions, begins with a chapter on polynomials. Chapter 2 is devoted to various types of field extensions, including finite, finitely generated, algebraic and normal. Chapter 3 takes a close look at the issue of separability. In my classes, I generally cover only Sections 3.1 to 3.4 (on perfect fields). Chapter 4 is devoted to algebraic independence, starting with the general notion of a dependence relation and concluding with Luroth's theorem on intermediate fields of a simple transcendental extension. Part 2 of the book is entitled Galois Theory. Chapter 5 examines Galois theory from an historical perspective, discussing the contributions from Lagrange, Vandermonde, Gauss, Newton, and others that led to the development of the theory. I have also included a very brief look at the very brief life of Galois himself. Chapter 6 begins with the notion of a Galois correspondence between two partially ordered sets, and then specializes to the Galois correspondence of a field extension, concluding with a brief discussion of the Krull topology. In Chapter 7, we discuss the Galois theory of equations. In Chapter 8, we view a field extension of as a vector space over . Chapter 9 and Chapter 10 are devoted to finite fields, although this material can be omitted in order to reach the topic of solvability by radicals more quickly. Mo bius inversion is used in a few places, so an appendix has been included on this subject. viii Preface Part 3 of the book is entitled The Theory of Binomials. Chapter 11 covers the roots of unity and Wedderburn's theorem on finite division rings. We also briefly discuss the question of whether a given group is the Galois group of a field extension. In Chapter 12, we characterize cyclic extensions and splitting fields of binomials when the base field contains appropriate roots of unity. Chapter 13 is devoted to the question of solvability of a polynomial equation by radicals. (This chapter might make a convenient ending place in a graduate course.) In Chapter 14, we determine conditions that characterize the irreducibility of a binomial and describe the Galois group of a binomial. Chapter 15 briefly describes the theory of families of binomialsthe socalled Kummer theory. Sections marked with an asterisk may be skipped without loss of continuity. Changes for the Second Edition
Let me begin by thanking the readers of the first edition for their many helpful comments and suggestions. For the second edition, I have gone over the entire book, and rewritten most of it, including the exercises. I believe the book has benefited significantly from a class testing at the beginning graduate level and at a more advanced graduate level. I have also rearranged the chapters on separability and algebraic independence, feeling that the former is more important when time is of the essence. In my course, I generally touch only very lightly (or skip altogether) the chapter on algebraic independence, simply because of time constraints. As mentioned earlier, as several readers have requested, I have added a chapter on Galois theory from an historical perspective. A few additional topics are sprinkled throughout, such as a proof of the Fundamental Theorem of Algebra, a discussion of casus irreducibilis, Berlekamp's algorithm for factoring polynomials over and natural and accessory irrationalities. Thanks
I would like to thank my students Phong Le, Sunil Chetty, Timothy Choi and Josh Chan, who attended lectures on essentially the entire book and offered many helpful suggestions. I would also like to thank my editor, Mark Spencer, who puts up with my many requests and is most amiable. Contents Preface....................................................................................................vii Contents...................................................................................................ix 0 Preliminaries...................................................................................1
0.1 Lattices..................................................................................................1 0.2 Groups.................................................................................................. 2 0.3 The Symmetric Group........................................................................ 10 0.4 Rings...................................................................................................10 0.5 Integral Domains................................................................................ 14 0.6 Unique Factorization Domains........................................................... 16 0.7 Principal Ideal Domains..................................................................... 16 0.8 Euclidean Domains.............................................................................17 0.9 Tensor Products.................................................................................. 17 Exercises...................................................................................................19 Part IField Extensions 1 Polynomials................................................................................... 23
1.1 Polynomials over a Ring.....................................................................23 1.2 Primitive Polynomials and Irreducibility............................................24 1.3 The Division Algorithm and Its Consequences.................................. 27 1.4 Splitting Fields....................................................................................32 1.5 The Minimal Polynomial.................................................................... 32 1.6 Multiple Roots.................................................................................... 33 1.7 Testing for Irreducibility.....................................................................35 Exercises...................................................................................................38 2 Field Extensions............................................................................41
2.1 The Lattice of Subfields of a Field..................................................... 41 2.2 Types of Field Extensions.................................................................. 42 2.3 Finitely Generated Extensions............................................................ 46 2.4 Simple Extensions.............................................................................. 47 2.5 Finite Extensions................................................................................ 53 2.6 Algebraic Extensions.......................................................................... 54 x Contents 2.7 Algebraic Closures............................................................................. 56 2.8 Embeddings and Their Extensions..................................................... 58 2.9 Splitting Fields and Normal Extensions............................................. 63 Exercises...................................................................................................66 3 Embeddings and Separability..................................................... 73
3.1 Recap and a Useful Lemma................................................................ 73 3.2 The Number of Extensions: Separable Degree...................................75 3.3 Separable Extensions.......................................................................... 77 3.4 Perfect Fields...................................................................................... 84 3.5 Pure Inseparability.............................................................................. 85 *3.6 Separable and Purely Inseparable Closures...................................... 88 Exercises...................................................................................................91 4 Algebraic Independence...............................................................93
4.1 Dependence Relations........................................................................ 93 4.2 Algebraic Dependence........................................................................96 4.3 Transcendence Bases........................................................................ 100 *4.4 Simple Transcendental Extensions................................................. 105 Exercises.................................................................................................108 Part IIGalois Theory 5 Galois Theory I: An Historical Perspective............................. 113
5.1 The Quadratic Equation....................................................................113 5.2 The Cubic and Quartic Equations.....................................................114 5.3 HigherDegree Equations................................................................. 116 5.4 Newton's Contribution: Symmetric Polynomials..............................117 5.5 Vandermonde....................................................................................119 5.6 Lagrange........................................................................................... 121 5.7 Gauss................................................................................................ 124 5.8 Back to Lagrange..............................................................................128 5.9 Galois................................................................................................130 5.10 A Very Brief Look at the Life of Galois.........................................135 6 Galois Theory II: The Theory................................................... 137
6.1 Galois Connections...........................................................................137 6.2 The Galois Correspondence..............................................................143 6.3 Who's Closed?.................................................................................. 148 6.4 Normal Subgroups and Normal Extensions......................................154 6.5 More on Galois Groups.................................................................... 159 6.6 Abelian and Cyclic Extensions......................................................... 164 *6.7 Linear Disjointness......................................................................... 165 Exercises.................................................................................................168 7 Galois Theory III: The Galois Group of a Polynomial........... 173
7.1 The Galois Group of a Polynomial................................................... 173 7.2 Symmetric Polynomials.................................................................... 174 7.3 The Fundamental Theorem of Algebra.............................................179 Contents xi 7.4 The Discriminant of a Polynomial....................................................180 7.5 The Galois Groups of Some SmallDegree Polynomials..................182 Exercises.................................................................................................193 8 A Field Extension as a Vector Space........................................ 197
8.1 The Norm and the Trace................................................................... 197 *8.2 Characterizing Bases...................................................................... 202 *8.3 The Normal Basis Theorem............................................................206 Exercises.................................................................................................208 9 Finite Fields I: Basic Properties................................................ 211
9.1 Finite Fields Redux...........................................................................211 9.2 Finite Fields as Splitting Fields........................................................ 212 9.3 The Subfields of a Finite Field......................................................... 213 9.4 The Multiplicative Structure of a Finite Field.................................. 214 9.5 The Galois Group of a Finite Field...................................................215 9.6 Irreducible Polynomials over Finite Fields.......................................215 *9.7 Normal Bases..................................................................................218 *9.8 The Algebraic Closure of a Finite Field......................................... 219 Exercises.................................................................................................223 10 Finite Fields II: Additional Properties..................................... 225
10.1 Finite Field Arithmetic................................................................... 225 *10.2 The Number of Irreducible Polynomials...................................... 232 *10.3 Polynomial Functions................................................................... 234 *10.4 Linearized Polynomials................................................................ 236 Exercises.................................................................................................238 11 The Roots of Unity......................................................................239
11.1 Roots of Unity................................................................................ 239 11.2 Cyclotomic Extensions................................................................... 241 *11.3 Normal Bases and Roots of Unity................................................ 250 *11.4 Wedderburn's Theorem.................................................................251 *11.5 Realizing Groups as Galois Groups..............................................253 Exercises.................................................................................................257 12 Cyclic Extensions........................................................................261
12.1 Cyclic Extensions........................................................................... 261 12.2 Extensions of Degree Char .......................................................265 Exercises.................................................................................................266 13 Solvable Extensions.................................................................... 269
13.1 Solvable Groups............................................................................. 269 13.2 Solvable Extensions........................................................................270 13.3 Radical Extensions......................................................................... 273 13.4 Solvability by Radicals................................................................... 274 13.5 Solvable Equivalent to Solvable by Radicals................................. 276 13.6 Natural and Accessory Irrationalities............................................. 278 13.7 Polynomial Equations.....................................................................280 xii Contents Exercises.................................................................................................282 Part IIIThe Theory of Binomials 14 Binomials.....................................................................................289
14.1 Irreducibility................................................................................... 289 14.2 The Galois Group of a Binomial.................................................... 296 *14.3 The Independence of Irrational Numbers..................................... 304 Exercises.................................................................................................307 15 Families of Binomials................................................................. 309
15.1 The Splitting Field.......................................................................... 309 15.2 Dual Groups and Pairings...............................................................310 15.3 Kummer Theory..............................................................................312 Exercises.................................................................................................316 Appendix: Mbius Inversion.............................................................. 319
Partially Ordered Sets............................................................................. 319 The Incidence Algebra of a Partially Ordered Set.................................. 320 Classical Mo bius Inversion.....................................................................324 Multiplicative Version of Mo bius Inversion.......................................... 325 References............................................................................................ 327 Index..................................................................................................... 329 Chapter 0 Preliminaries The purpose of this chapter is to review some basic facts that will be needed in the book. The discussion is not intended to be complete, nor are all proofs supplied. We suggest that the reader quickly skim this chapter (or skip it altogether) and use it as a reference if needed. 0.1 Lattices
Definition A partially ordered set (or poset) is a nonempty set , together with a binary relation on satisfying the following properties. For all , , , 1) (reflexivity) 2) (antisymmetry) , 3) (transitivity) , If, in addition, , then is said to be totally ordered. or Any subset of a poset is also a poset under the restriction of the relation defined on . A totally ordered subset of a poset is called a chain. If and for all then is called an upper bound for . A least upper bound for , denoted by lub or , is an upper bound that is less than or equal to any other upper bound. Similar statements hold for lower bounds and greatest lower bounds, the latter denoted by glb , or . A maximal element in a poset is an element such that implies .A minimal element in a poset is an element such that implies 2 Field Theory . A top element is an element with the property that for all . Similarly, a bottom element is an element with the property that for all . Zorn's lemma says that if every chain in a poset has an upper bound in then has a maximal element. Definition A lattice is a poset in which every pair of elements , has a least upper bound, or join, denoted by and a greatest lower bound, or meet, denoted by . If every nonempty subset of has a join and a meet then is called a complete lattice. Note that any nonempty complete lattice has a greatest element, denoted by and a smallest element, denoted by . Definition A sublattice of a lattice meet and join operation of . is a subset of that is closed under the It is important to note that a subset of a lattice can be a lattice under the same order relation and yet not be a sublattice of . As an example, consider the coll of all subgroups of a group , ordered by inclusion. Then is a subset of the power set , which is a lattice under union and intersection. But is not a sublattice of since the union of two subgroups need not be a subgroup. On the other hand, is a lattice in its own right under set inclusion, where the meet of two subgroups is their intersection and the join is the smallest subgroup of containing and . In a complete lattice , joins can be defined in terms of meets, since is the meet of all upper bounds of . The fact that ensures that has at least one upper bound, so that the meet is not an empty one. The following theorem exploits this idea to give conditions under which a subset of a complete lattice is itself a complete lattice. Theorem 0.1.1 Let be a complete lattice. If has the properties 1) 2) (Closed under arbitrary meets) , then is a complete lattice under the same meet. Proof. Let . Then by assumption. Let be the set of all upper bounds of that lie in . Since , we have . Hence, and is . Thus, is a complete lattice. (Note that need not be a sublattice of since need not equal the meet of all upper bounds of in .) 0.2 Groups
Definition A group is a nonempty set , together with a binary operation on , that is, a map , denoted by juxtaposition, with the following properties: Preliminaries 3 1) (Associativity) for all , , 2) (Identity) There exists an element for which 3) (Inverses) For each , there is an element . A group is abelian, or commutative, if , for all , The identity element is often denoted by . When operation is often denoted by and the identity by . for all for which . is abelian, the group Subgroups
Definition A subgroup of a group is a subset of that is a group in its own right, using the restriction of the operation defined on . We denote the fact that is a subgroup of by writing . If is a group and , then the set of all powers of is a subgroup of , called the cyclic subgroup generated by . A group is cyclic if it has the form , for some . In this case, we say that generates . Let be a group. Since is a subgroup of itself and since the intersection of subgroups of is a subgroup of , Theorem 0.1.1 implies that the set of subgroups of forms a complete lattice, where and is the smallest subgroup of containing both and . If and are subgroups of , it does not follow that the set product is a subgroup of . It is not hard to show that when . The center of is the set is a subgroup of precisely for all of all elements of that commute with every element of . Orders and Exponents
A group is finite if it contains only a finite number of elements. The cardinality of a finite group is called its order and is denoted by or . If , and if for some integer , we say that is an exponent of . The smallest positive exponent for is called the order of and is denoted by . An integer for which for all is called an 4 Field Theory exponent of . (Note: Some authors use the term exponent of smallest positive exponent of .) to refer to the Theorem 0.2.1 Let be a group and let . Then is an exponent of if and only if is a multiple of . Similarly, the exponents of are precisely the multiples of the smallest positive exponent of . We next characterize the smallest positive exponent for finite abelian groups. Theorem 0.2.2 Let be a finite abelian group. 1) (Maximum order equals minimum exponent) If is the maximum order of all elements in then for all . Thus, the smallest positive exponent of is equal to the maximum order of all elements of . 2) The smallest positive exponent of is equal to if and only if is cyclic. Cosets and Lagrange's Theorem
Let . We may define an equivalence relation on by saying that if (or equivalently . The equivalence classes are the left cosets of in . Thus, the distinct left cosets of form a partition of . Similarly, the distinct right cosets form a partition of . It is not hard to see that all cosets of have the same cardinality and that there is the same number of left cosets of in as right cosets. (This is easy when is finite. Otherwise, consider the map .) Definition The index of in , denoted by set of all distinct left cosets of in . If . , is the cardinality of the is finite then Theorem 0.2.3 Let be a finite group. 1) (Lagrange) The order of any subgroup of divides the order of . 2) The order of any element of divides the order of . 3) (Converse of Lagrange's Theorem for Finite Abelian Groups) If is a finite abelian group and if A then has a subgroup of order . Normal Subgroups
If and are subsets of a group , then the set product is defined by Theorem 0.2.4 Let . The following are equivalent 1) The set product of any two cosets is a coset. 2) If , then Preliminaries 5 3) Any right coset of for which 4) If , then is also a left coset, that is, for any . there is a 5) for all , . , written , if any of the Definition A subgroup of is normal in equivalent conditions in Theorem 0.2.4 holds. Definition A group and . is simple if it has no normal subgroups other than Here are some normal subgroups. Theorem 0.2.5 1) The center is a normal subgroup of . 2) Any subgroup of a group with is normal. 3) If is a finite group and if is the smallest prime dividing subgroup of index is normal in , then any With respect to the last statement in the previous theorem, it makes some intuitive sense that if a subgroup of a finite group is extremely large, then it may be normal, since there is not much room for conjugates. This is true in the most extreme case. Namely, the largest possible proper subgroup of has index equal to the smallest prime number dividing . This subgroup, if it exists, is normal. If , then we have the set product formula It is not hard to see that this makes the quotient into a group, called the quotient group of in . The order of is called the index of in and is denoted by . Theorem 0.2.6 If is a group and is a collection of normal subgroups of then and are normal subgroups of . Hence, the collection of normal subgroups of is a complete sublattice of the complete lattice of all subgroups of . If then there is always an intermediate subgroup for which , in fact, is such an intermediate subgroup. The largest such subgroup is called the normalizer of in . It is 6 Field Theory Euler's Formula
We will denote a greatest common divisor of and by or gcd . If , then and are relatively prime. The Euler phi function is defined by letting be the number of positive integers less than or equal to that are relatively prime to . Two integers and are congruent modulo , written is divisible by . Let denote the ring of integers addition and multiplication modulo . Theorem 0.2.7 (Properties of Euler's phi function) 1) The Euler phi function is multiplicative, that is, if prime, then mod , if 0 under and are relatively 2) If is a prime and then These two properties completely determine . Since the set multiplication modulo , it follows that Theorem 0.2.8 (Euler's Theorem) If mod Corollary 0.2.9 (Fermat's Theorem) If then is a prime not dividing the integer , mod is a group of order is an exponent for . and , then under Cyclic Groups
Theorem 0.2.10 1) Every group of prime order is cyclic. 2) Every subgroup of a cyclic group is cyclic. 3) A finite abelian group is cyclic if and only if its smallest positive exponent is equal to . The following theorem contains some key results about finite cyclic groups. Theorem 0.2.11 Let be a cyclic group of order . Preliminaries 7 1) For , In particular, 2) If , then generates if and only if , where . Thus the elements of of order are the elements of the form , where and is relatively prime to . 3) For each , the group has exactly one subgroup of order and elements of order , all of which lie in . 4) (Subgroup structure charactertizes property of being cyclic) If a finite group of order has the property that it has at most one subgroup of each order , then is cyclic. Counting the elements in a cyclic group of order corollary. Corollary 0.2.12 For any positive integer , gives the following Homomorphisms
Definition Let and homomorphism if be groups. A map is called a group A surjective homomorphism is an epimorphism, an injective homomorphism is a monomorphism and a bijective homomorphism is an isomorphism. If is an isomorphism, we say that and are isomorphic and write . If is a homomorphism then homomorphism , ker is a normal subgroup of . Conversely, any normal subgroup of is the kernel of a homomorphism. For we may define the natural projection by . This is easily seen to be an epimorphism with kernel . and . The kernel of a 8 Field Theory Let be a function from a set to a set . Let and be the power sets of and , respectively. We define the induced map by and the induced inverse map by . (It is customary to denote the induced maps by the same notation as the original map.) Note that is surjective if and only if its induced map is surjective, and this holds if and only if the induced inverse map is injective. A similar statement holds with the words surjective and injective reversed. Theorem 0.2.13 Let 1) a) If then b) If is surjective and 2) a) If then b) If then be a group homomorphism. . then . . . Theorem 0.2.14 Let be a group. 1) (First Isomorphism Theorem) Let be a group homomorphism with kernel . Then and the map im defined by is an isomorphism. Hence im . In particular, is injective if and only if ker . 2) (Second Isomorphism Theorem) If and then and 3) (Third Isomorphism Theorem) If then and with and normal in Hence Theorem 0.2.15 Let and . be groups and let . Then Theorem 0.2.16 (The Correspondence Theorem) Let natural projection . Thus, for any I 1) The induced maps and the lattice of subgroups of . and let , be the define a onetoone correspondence between containing and the lattice of subgroups of Preliminaries 9 2) preserves index, that is, for any , we have 3) preserves normality, that is, if , in which case then if and only if Theorem 0.2.17 1) An abelian group is simple if and only if it is finite and has prime order. 2) If is a maximal subgroup of , that is, and if then or , and if is normal then is cyclic of prime order. Sylow Subgroups
Definition If is a prime, then a group is called a group if every element of has order a power of . A Sylow subgroup of is a maximal subgroup of . Theorem 0.2.18 (Properties of groups) 1) A finite group is a group if and only if for some . 2) If is a finite group, then the center of is nontrivial. 3) If , prime, then is abelian. 4) If is a proper subgroup of , then is also a proper subgroup of its normalizer . 5) If is a maximal subgroup of then is normal and has index . For finite groups, if then general, but we do have the following. . The converse does not hold in Theorem 0.2.19 Let be a finite group. 1) (Cauchy's Theorem) If is divisible by a prime then contains an element of order . 2) (Partial converse of Lagrange's theorem) If is a prime and , then for any Sylow subgroup of , there is a subgroup of , normal in and of order . Here is the famous result on maximal subgroups of a finite group. Theorem 0.2.20 (Sylow's Theorem) Let have order where . 1) All Sylow subgroups of have order . 2) All Sylow subgroups are conjugate (and hence isomorphic). 3) The number of Sylow subgroups of divides and is congruent to mod . 4) Any subgroup of is contained is a Sylow subgroup of . 10 Field Theory 0.3 The Symmetric Group
Definition The symmetric group on the set is the group of all permutations of , under composition of maps. A transposition is a permutation that interchanges two distinct elements of and leaves all other elements fixed. The alternating group is the subgroup of consisting of all even permutations, that is, all permutations that can be written as a product of an even number of transpositions. Theorem 0.3.1 1) The order of is . 2) The order of is 2. Thus, and A 3) is the only subgroup of of index . 4) is simple (no nontrivial normal subgroups for A subgroup . of is transitive if for any . . for which there is a Theorem 0.3.2 If multiple of . is a transitive subgroup of then the order is a 0.4 Rings
Definition A ring is a nonempty set , together with two binary operations on , called addition (denoted by ), and multiplication (denoted by juxtaposition), satisfying the following properties. 1) is an abelian group under the operation . 2) (Associativity) for all . 3) (Distributivity) For all , and Definition Let be a ring. 1) is called a ring with identity if there exists an element for which , for all . In a ring with identity, an element is called a unit if it has a multiplicative inverse in , that is, if there exists a such that . 2) is called a commutative ring if multiplication is commutative, that is, if for all . 3) A zero divisor in a commutative ring is a nonzero element such that for some 0. A commutative ring with identity is called an integral domain if contains no zero divisors. 4) A ring with identity is called a field if the nonzero elements of form an abelian group under multiplication. It is not hard to see that the set of all units in a ring with identity forms a group under multiplication. We shall have occasion to use the following example. Preliminaries 11 Example 0.4.1 Let be the ring of integers modulo . Then is a unit in if and only if . This follows from the fact that if and only if there exist integers and such that , that is, if and only if mod . The set of units of , denoted by , is a group under multiplication. Definition A subring of a ring is a nonempty subset of its own right, using the same operations as defined on . that is a ring in Definition A subfield of a field is a nonempty subset of that is a field in its own right, using the same operations as defined on . In this case, we say that is an extension of and write or . Definition Let and for all , , be rings. A function and An injective homomorphism is a monomorphism or an embedding, a surjective homomorphism is an epimorphism and a bijective homomorphism is an isomorphism. A homomorphism from into itself is an endomorphism and an isomorphism from onto itself is an automorphism. is a homomorphism if, Ideals
Definition A nonempty subset 1) implies 2) , implies of a ring . and is called an ideal if it satisfies . If is a nonempty subset of a ring , then the ideal generated by is defined to be the smallest ideal of containing . If is a commutative ring with identity, and if , then the ideal generated by is the set Any ideal of the form Definition If is called a principal ideal. is a homomorphism, then ker is an ideal of If . is an ideal in then for each , we can form the coset is a ring and It is easy to see that if and only if , and that any two 12 Field Theory cosets and are either disjoint or identical. The collection of all (distinct) cosets is a ring itself, with addition and multiplication defined by and The ring of cosets of is called a factor ring and is denoted by . Isomorphism theorems similar to those for groups also hold for rings. Here is the first isomorphism theorem. Theorem 0.4.1 (The First Isomorphism Theorem) Let be a ring. Let be a ring homomorphism with kernel . Then is an ideal of and the map im defined by is an isomorphism. Hence im . In particular, is injective if and only if ker . Definition An ideal of a ring is maximal if and if whenever for any ideal , then or . An ideal is prime if and if implies or . It is not hard to see that a maximal ideal in a commutative ring with identity is prime. This also follows from the next theorem. Theorem 0.4.2 Let be a commutative ring with identity and let of . 1) is a field if and only if is maximal. 2) is an integral domain if and only if is prime. be an ideal Theorem 0.4.3 Any commutative ring with identity contains a maximal ideal. Proof. Since is not the zero ring, the ideal is a proper ideal of . Hence, the set of all proper ideals of is nonempty. If is a chain of proper ideals in then the union is also an ideal. Furthermore, if is not proper, then and so , for some , which implies that is not proper. Hence, . Thus, any chain in has an upper bound in and so Zorn's lemma implies that has a maximal element. This shows that has a maximal ideal. Preliminaries 13 The Characteristic of a Ring
Let be a ring and let . For any positive integer , we define terms and for any negative integer , we set . The characteristic char of a ring is the smallest positive integer for which (or equivalently, for all ), should such an integer exist. If it does not, we say that has characteristic . If char then contains a copy of the integers , in the form . If char , then contains a copy of . Theorem 0.4.4 The characteristic of an integral domain is either or a prime. In particular, a finite field has prime characteristic. Proof. If char is not and if , where and are positive integers, then and so one of or is equal to . But since is the smallest such positive integer, it follows that either or . Hence, is prime. If is a field, the intersection of all of its subfields is the smallest subfield of and is referred to as the prime subfield of . Theorem 0.4.5 Let be a field. If char isomorphic to the rational numbers . If char of is isomorphic to . Proof. If char , consider the map , the prime subfield of is is prime, the prime field defined by This is easily seen to be a ring homomorphism. For example Now, if and only if in , and since char , we see 14 Field Theory that and so is a monomorphism. Thus, the subfield is isomorphic to . Clearly, any subfield of must contain the elements , where and therefore also the elements and so is the prime subfield of . Now suppose that char is a prime. The map defined by is a ring homomorphism and is also injective since . Hence, is a subfield of isomorphic to . Since any subfield of must contain , this is the prime subfield of . The following result is of considerable importance for the study of fields of nonzero characteristic. Theorem 0.4.6 Let characteristic . Then be a commutative ring with identity of prime Proof. Since the binomial formula holds in any commutative ring with identity, we have where But for 0 therefore reduces to , and so in . The binomial formula Repeated use of this formula gives proved similarly. . The second formula is These formulas are very significant. They say that the Frobenius map is a surjective ring homomorphism. When is a field of characteristic , then is an isomorphism and . 0.5 Integral Domains
Theorem 0.5.1 Let be an integral domain. Let , . 1) We say that divides and write if for some are nonunits and then properly divides . a) A unit divides every element of . b) if and only if . c) properly if and only if . . If and Preliminaries 15 2) If for some unit u then and are associates and we write . a) if and only if and . b) if and only if . 3) A nonzero element is irreducible if is not a unit and if has no proper divisors. Thus, a nonunit is irreducible if and only if implies that either or is a unit. 4) A nonzero element is prime if is not a unit and whenever then or . a) Every prime element is irreducible. b) is prime if and only if is a nonzero prime ideal. 5) Let , . An element is called a greatest common divisor (gcd) of and , written or gcd , if and and if whenever , then . If gcd is a unit, we say that and are relatively prime. The greatest common divisor of two elements, if it exists, is unique up to associate. Theorem 0.5.2 An integral domain is a field if and only if it has no ideals other than the zero ideal and itself. Any nonzero homomorphism of fields is a monomorphism. Theorem 0.5.3 Every finite integral domain is a field. Field of Quotients
If is an integral domain, we may form the set , where if and only if on in the "obvious way" , 0 . We define addition and multiplication It is easy to see that these operations are welldefined and that is actually a field, called the field of quotients of the integral domain . It is the smallest field containing (actually, an isomorphic copy of ), in the sense that if is a field and then . The following fact will prove useful. Theorem 0.5.4 Let be an integral domain with field of quotients . Then any monomorphism from into a field has a unique extension to a monomorphism . Proof. Define , which makes sense since implies . One can easily show that is welldefined. Since if and only if , which in turn holds if and only if , we see that is injective. Uniqueness is clear since ( restricted to ) uniquely determines on . 16 Field Theory 0.6 Unique Factorization Domains
Definition An integral domain is a unique factorization domain (ufd) if 1) Any nonunit can be written as a product where is irreducible for all . We refer to this as the factorization property for . 2) This factorization is essentially unique in the sense that if are two factorizations into irreducible elements then and there is some permutation for which for all . If is not irreducible, then where and are nonunits. Evidently, we may continue to factor as long as at least one factor is not irreducible. An integral domain has the factorization property precisely when this factoring process always stops after a finite number of steps. Actually, the uniqueness part of the definition of a ufd is equivalent to some very important properties. Theorem 0.6.1 Let be an integral domain for which the factorization property holds. The following conditions are equivalent and therefore imply that is a unique factorization domain. 1) Factorization in is essentially unique. 2) Every irreducible element of is prime. 3) Any two elements of , not both zero, have a greatest common divisor. Corollary 0.6.2 In a unique factorization domain, the concepts of prime and irreducible are equivalent. 0.7 Principal Ideal Domains
Definition An integral domain every ideal of is principal. is called a principal ideal domain (pid) if Theorem 0.7.1 Every principal ideal domain is a unique factorization domain. We remark that the ring is a ufd (as we prove in Chapter 1) but not a pid (the ideal is not principal) and so the converse of the previous theorem is not true. Theorem 0.7.2 Let be a principal ideal domain and let be an ideal of 1) is maximal if and only if where is irreducible. 2) is prime if and only if or is maximal. 3) The following are equivalent: a) is a field b) is an integral domain c) is irreducible . Preliminaries 17 d) is prime. 0.8 Euclidean Domains
Roughly speaking, a Euclidean domain is an integral domain in which we can perform "division with remainder." Definition An integral domain is a Euclidean domain if there is a function with the property that given any , , , there exist satisfying where or . Theorem 0.8.1 A Euclidean domain is a principal ideal domain (and hence also a unique factorization domain). Proof. Let be an ideal in the Euclidean domain and let be minimal with respect to the value of . Thus, for all . If then where leaving or and . But . Hence, and so the latter is not possible, . Theorem 0.8.2 If is a field, then is a Euclidean domain with deg . Hence is also a principal ideal domain and a unique factorization domain. Proof. This follows from ordinary division of polynomials; to wit, if , then there exist such that where deg deg . 0.9 Tensor Products
Tensor products are used only in the optional Section 5.6, on linear disjointness. Definition Let , and be vector spaces over a field . A function is bilinear if it is linear in both variables separately, that is, if and The set of all bilinear functions from to is denoted by bilinear function , with values in the base field bilinear form on . .A , is called a 18 Field Theory Example 0.9.1 1) A real inner product , is a bilinear form on 2) If is an algebra, the product map defined by is bilinear. . We will denote the set of all linear transformations from to by . There are many definitions of the tensor product. We choose a universal definition. Theorem 0.9.1 Let and exists a unique vector space the following property. If to a vector space over for which be vector spaces over the same field . There and bilinear map with is any bilinear function from , then there is a unique linear transformation This theorem says that to each bilinear function , there corresponds a unique linear function , through which can be factored (that is, . The vector space , whose existence is guaranteed by the previous theorem, is called the tensor product of and over . We denote the image of under the map by u . If im , is the image of the tensor map then the uniqueness statement in the theorem implies that spans . Hence, every element of is a finite sum of elements of the form
finite We establish a few basic properties of the tensor product. Theorem 0.9.2 If then is linearly for all Proof. Consider the dual vectors to the vectors , where For linear functionals , we define a bilinear form
j , independent and . by Since there exists a unique linear functional , we have for which Preliminaries 19 j Since the 's are arbitrary, we deduce that Corollary 0.9.3 If Theorem 0.9.4 Let be a basis for . Then Proof. To see that the and , then for all . . and let is a basis for is linearly independent, suppose that
,j j be a basis for . This can be written
, Theorem 0.9.2 implies that
, for all , and hence , . Since for all and . To see that spans , and , we have , let Since any vector in spans . is a finite sum of vectors , we deduce that Corollary 0.9.5 For finite dimensional vector spaces, dim dim dim Exercises
1. 2. The relation of being associates in an integral domain is an equivalence relation. Prove that the characteristic of an integral domain is either or a prime, and that a finite field has prime characteristic. 20 Field Theory If char , the prime subfield of is isomorphic to the rational numbers . If char is prime, the prime field of is isomorphic to . 4. If show that and must have the same characteristic. 5. Let be a field of characteristic . The Frobenius map defined by is a homomorphism. Show that . What if is a finite field? 6. Consider the polynomial ring where 2 . Show that the factorization process need not stop in this ring. 7. Let . Show that this integral domain is not a unique factorization domain by showing that has essentially two different factorizations in . Show also that the irreducible element is not prime. 8. Let be a pid. Then an ideal of is maximal if and only if where is irreducible. Also, is a field if and only if is irreducible. 9. Prove that and are both prime ideals in and that is properly contained in . 10. Describe the divisor chain condition in terms of principal ideals. 3. Part IField Extensions Chapter 1 Polynomials In this chapter, we discuss properties of polynomials that will be needed in the sequel. Since we assume that the reader is familiar with the basic properties of polynomials, some of the present material may constitute a review. 1.1 Polynomials over a Ring
We will be concerned in this book mainly with polynomials over a field , but it is useful to make a few remarks about polynomials over a ring as well, especially since many polynomials encountered in practice are defined over the integers. Let denote the ring of polynomials in the single variable over . If where a and 0 then is called the degree of , written deg or deg and is called the leading coefficient of . A polynomial is monic if its leading coefficient is . The degree of the zero polynomial is defined to be . If is a ring, the units of are the units of , since no polynomial of positive degree can have an inverse in . Note that the units in are the units in . In general, if and if that sends to is a polynomial over a ring is a ring homomorphism, then we denote the polynomial by or by and the function by , that is, We may refer to as the extension of a ring homomorphism. to . It is easy to see that is also One of the most useful examples of ring homomorphisms in this context is the projection maps , where is a prime in , defined by . It is not hard to see that is a surjective ring 24 Field Theory homomorphism, and that is an integral domain. The maps referred to as localization maps. Note that the units of are the units of . are also Definition Let be a ring. A nonzero polynomial is irreducible over if is not a unit and whenever for , then one of and is a unit in . A polynomial that is not irreducible is said to be reducible. We can simplify this definition for polynomials over a field. A polynomial over a field is irreducible if and only if it has positive degree and cannot be factored into the product of two polynomials of positive degree. Many important properties that a ring polynomials . may possess carry over to the ring of Theorem 1.1.1 Let be a ring. 1) If is an integral domain, then so is 2) If is a unique factorization domain, then so is . 3) If is a principal ideal domain, need not be a principal ideal domain. 4) If is a field, then is a principal ideal domain. Proof. For part 3), the ring of integers is a principal ideal domain, but is not, since the ideal is not principal. 1.2 Primitive Polynomials and Irreducibility
We now consider polynomials over a unique factorization domain. Content and Primitivity
If is a polynomial over the integers, it is often useful to factor out the positive greatest common divisor of the coefficients, so that the remaining coefficients are relatively prime. For polynomials over an arbitrary unique factorization domain, the greatest common divisor is not unique and there is no way to single one out in general. Definition Let where is a unique factorization domain. Any greatest common divisor of the coefficients of is called a content of . A polynomial with content is said to be primitive. Let denote the set of all contents of . Thus, is the set of all associates of any one of its elements. For this reason, one often speaks of "the" content of a polynomial. A content of can be obtained by factoring each coefficient of into a product of powers of distinct primes and then taking the product of each prime that appears in any of these factorizations, raised to the smallest power to which appears in all of the factorizations. Polynomials 25 There is no reason why we cannot apply this same procedure to a polynomial over , the field of quotients of . If , then each coefficient of can be written as a product of integral powers of distinct primes. Definition Let quotients of a unique factorization domain . Let of the distinct primes dividing any coefficient of can be written in the form , where is the field of be a complete list . Then each coefficient where exponent of . Let min be the smallest among the factorizations of the coefficients of . The element is a content of , and so is any element , where is a unit in . The set of all contents of is denoted by . A polynomial is primitive if . Note the following simple facts about content. Lemma 1.2.1 For any and It follows that 1) is a content of if and only if in . 2) If is primitive, then . 3) if and only if . , where is primitive We now come to a key result concerning primitive polynomials. Theorem 1.2.2 Let be a unique factorization domain, with field of quotients . 1) (Gauss's lemma) The product of primitive polynomials is primitive. 2) If then . 3) If a polynomial can be factored where is primitive and Proof. To prove Gauss's lemma, let not primitive. Then there exists a prime localization map . The condition then, in fact, . and suppose that is for which . Consider the is equivalent to , that is, 26 Field Theory and since is an integral domain, one of the factors must be , that is, one of or must be divisible by , and hence not primitive. To prove part 2), observe that if is a content of and is a content of then and , where and are primitive over . Hence, by Gauss's lemma, if is the set of units of , then As to part 3), we have and since , so is , whence . Irreducibility over and If , then it can also be thought of as a polynomial over . We would like to relate the irreducibility of over to its irreducibility over . Let us say that a factorization is over a set if and have coefficients in . The relationship between irreducibility over and over would be quite simple were it not for the presence of irreducible constants in , which are not irreducible over . To formulate a clear description of the situation, let us make the following nonstandard (not found in other books) definition. We say that a factorization of the form , where deg and deg , is a degreewise factorization of and that is degreewise reducible. Now, if is a degreewise factorization over , then it is also a degreewise factorization over . Conversely, if this is a degreewise factorization over , then we can move the content of to the other factor and write where is primitive. Theorem 1.1.1 implies that is also in and so this is a degreewise factorization of over . Thus, has a degreewise factorization over if and only if it has a degreewise factorization over . Note also that the corresponding factors in the two factorizations have the same degree. Polynomials 27 It follows that is irreducible over if and only if it is degreewise irreducible over . But degreewise irreducibility over a field is the only kind of irreducibility. Theorem 1.2.3 Let be a unique factorization domain, with field of quotients . Let . 1) is degreewise irreducible over if and only if it is irreducible over . 2) If is primitive, then it is irreducible over if and only if it is irreducible over . 1.3 The Division Algorithm and its Consequences
The familiar division algorithm for polynomials over a field can be easily extended to polynomials over a commutative ring with identity, provided that we divide only by polynomials with leading coefficient a unit. We leave proof of the following to the reader. Theorem 1.3.1 (Division algorithm) Let be a commutative ring with identity. Let have an invertible leading coefficient (which happens if is monic, for example). Then for any , there exist unique such that where deg deg . This theorem has some very important immediate consequences. Dividing by , where gives where over . . Hence, is a root of if and only if is a factor of Corollary 1.3.2 Let Then is a root of be a commutative ring with identity and let if and only if is a factor of over . . Also, since the usual degree formula deg deg deg holds when is an integral domain, we get an immediate upper bound on the number of roots of a polynomial. Corollary 1.3.3 If is an integral domain, then a nonzero polynomial can have at most deg distinct roots in . 28 Field Theory Note that if example, in . is not an integral domain then the preceding result fails. For , the four elements and are roots of the polynomial From this, we get the following fundamental fact concerning finite multiplicative subgroups of a field. Corollary 1.3.4 Let be the multiplicative group of all nonzero elements of a field . If is a finite subgroup of , then is cyclic. In particular, if is a finite field then is cyclic. Proof. If , then every element of satisfies the polynomial . But cannot have an exponent , for then every one of the elements of would be a root of the polynomial , of degree less than . Hence, the smallest exponent of is the order of and Theorem 0.2.2 implies that is cyclic. Polynomials as Functions
In the customary way, a polynomial can be thought of as a function on . Of course, the zero polynomial is also the zero function. However, the converse is not true! For example, the nonzero polynomial in is the zero function on . This raises the question of how to decide, based on the zero set of a polynomial, when that polynomial must be the zero polynomial. If is an integral domain, then Corollary 1.3.3 ensures that if has degree at most but has more than zeros, then it must be the zero polynomial. The previous example shows that we cannot improve on this statement. It follows that if the zero set of is infinite, then must be the zero polynomial. We can make no such blanket statements in the context of finite rings, as the previous example illustrates. Now let us consider polynomials in more than one variable. We can no longer claim that if a polynomial has an infinite zero set, then it must be the zero polynomial. For example, the nonzero polynomial has the infinite zero set . It is not hard to prove by induction that if is infinite and is the zero function, that is, has zero set , then is the zero polynomial. We leave the details to the reader. Again, we cannot strengthen this to finite rings, as the polynomial in shows. However, we can improve upon this. There is a middle ground between "an infinite set of zeros" and "zero set equal to all of " that is sufficient to Polynomials 29 guarantee that is the zero polynomial. This middle ground is "an infinite subfield worth of zeros." Theorem 1.3.5 Let is infinite. If polynomial. Proof. Write be a polynomial over for all , then and let , where is the zero where . Let be a basis for as a vector space over
, . Then for , and so
, , Hence, the independence of the 's implies that the polynomial
, in that and , is the zero function on . As we have remarked, this implies for all and . Hence, for all is the zero polynomial. Common Divisors and Greatest Common Divisors
In defining the greatest common divisor of two polynomials, it is customary (in order to obtain uniqueness) to require that it be monic. Definition Let and be polynomials over . The greatest common divisor of and , denoted by or gcd , is the unique monic polynomial over for which 1) and . 2) If and and then . The existence of greatest common divisors is easily proved using the fact that is a principal ideal domain. Since the ideal 30 Field Theory is principal, we have , for some monic . Since , it follows that and . Moreover, since , there exist such that Hence, if gcd and . , then and so As to uniqueness, if and are both greatest common divisors of and then each divides the other and since they are both monic, we conclude that . Greatest Common Divisor Is Field Independent The definition of greatest common divisor seems at first to depend on the field , since all divisions are over . However, this is not the case. To see this, note that for any field , the ideal containing the coefficients of and is principal and so , where is the gcd with respect to the field . But if , then and so . This implies two things. First, because generates and second, because is the greatest common divisor of and in . Hence, . Thus, if is the smallest field containing the coefficients of and , then is the same polynomial as , for any field containing the coefficients of and . In other words, the gcd can be computed using any field containing the coefficients of and . This also shows that the gcd of and has coefficients in the field . Theorem 1.3.6 Let the coefficients of and . 1) The greatest common divisor the base field . 2) Hence, has coefficients in 3) There exist polynomials . Let of . such that be the smallest field containing and does not depend on the This result has a somewhat surprising corollary: If have a nonconstant common factor in any extension of , then gcd is nonconstant and so and have a nonconstant factor over every field containing the coefficients of and . Polynomials 31 Corollary 1.3.7 Let and let . Then and have a nonconstant common factor over if and only if they have a nonconstant common factor over . Now we can make sense of the notion that two polynomials are relatively prime without mentioning a specific field. Definition The polynomials and are relatively prime if they have no nonconstant common factors, that is, if gcd . In particular, and are relatively prime if and only if there exist polynomials and over the smallest field containing the coefficients of and for which Roots and Common Roots
It is a fundamental fact that every nonconstant polynomial root in some field. has a Theorem 1.3.8 Let be a field, and let be a nonconstant polynomial. Then there exists an extension of and an such that . Proof. We may assume that is irreducible. Consider the field The field is isomorphic to a subfield of . Under this identification, , by identifying is a root of in with . Thus, we have shown that can be embedded in a field in which (with its coefficients embedded as well) has a root. While this is not quite the statement of the theorem, it is possible to show that there is a "true" extension of that has a root of , using simple techniques from the next chapter. Repeated application of Theorem 1.3.8 gives the following corollary. Corollary 1.3.9 Let . There exists an extension of splits, that is, factors into linear factors. over which Corollary 1.3.10 Two polynomials have a nonconstant common factor over some extension of if and only if they have a common root over some extension of . Put another way, and are relatively prime if and only if they have no common roots in any extension . 32 Field Theory Since distinct irreducible polynomials are relatively prime, we get the following corollary. Corollary 1.3.11 If and are distinct irreducible polynomials over then they have no common roots in any extension of . 1.4 Splitting Fields
If a polynomial factors into linear factors in an extension field , that is, if , we say that splits in . Definition Let be family of polynomials over a field . A splitting field for is an extension field of with the following properties: 1) Each splits over , and thus has a full set of deg roots in 2) is the smallest field satisfying that contains the roots of each mentioned in part 1). Theorem 1.4.1 Every finite family of polynomials over a field has a splitting field. Proof. According to Corollary 1.3.9, there is an extension in which a given polynomial has a full set of roots . The smallest subfield of containing and these roots is a splitting field for . If is a finite family of polynomials, then a splitting field for is a splitting field for the product of the polynomials in . We will see in the next chapter that any family of polynomials has a splitting field. We will also see that any two splitting fields and for a family of polynomials over are isomorphic by an isomorphism that fixes each element of the base field . 1.5 The Minimal Polynomial
Let . An element is said to be algebraic over if is a root of some polynomial over . An element that is not algebraic over is said to be transcendental over . If is algebraic over , the set of all polynomials satisfied by is a nonzero ideal in and is therefore generated by a unique monic polynomial , called the minimal polynomial of over and denoted by , or min . The following theorem characterizes minimal polynomials in a variety of useful ways. Proof is left to the reader. Polynomials 33 Theorem 1.5.1 Let and let be algebraic over . Then among all polynomials in , the polynomial min is 1) the unique monic irreducible polynomial for which 2) the unique monic polynomial of smallest degree for which 3) the unique monic polynomial with the property that if and only if . In other words, min is the unique monic generator of the ideal . Definition Let . Then , are said to be conjugates over have the same minimal polynomial over . if they 1.6 Multiple Roots
Let us now explore the issue of multiple roots of a polynomial. Definition Let be a root of . The multiplicity of is the largest positive integer for which divides . If , then is a simple root and if , then is a multiple root of . Definition An irreducible polynomial is separable if it has no multiple roots in any extension of . An irreducible polynomial that is not separable is inseparable. We should make a comment about this definition. It is not standard. For example, Lang defines a polynomial to be separable if it has no multiple roots, saying nothing about irreducibility. Hence, is not separable under this definition. Jacobson defines a polynomial to be separable if its irreducible factors have no multiple roots. Hence, is separable under this definition. However, van der Waerden, who first proposed the term "separable", gave the definition we have adopted, which does require irreducibility. Hence, for us, the question of whether is separable is not applicable, since is not irreducible. The only inconvenience with this definition is that we cannot say that if is separable over , then it is also separable over an extension of . Instead we must say that the irreducible factors of are separable over . Although, as we will see, all irreducible polynomials over a field of characteristic zero or a finite field are separable, the concept of separability (that is, inseparability) plays a key role in the theory of more "unusual" fields. Theorem 1.6.1 A polynomial has no multiple roots if and only if and its derivative are relatively prime. Proof. Over a splitting field for , we have 34 Field Theory where the 's are distinct. It is easy to see that and nontrivial common factors over if and only if for all have no . Corollary 1.6.2 An irreducible polynomial is separable if and only if . Proof. Since deg deg and is irreducible, it follows that and are relatively prime if and only if . If char then following corollary. for any nonconstant . Thus, we get the Corollary 1.6.3 All irreducible polynomials over a field of characteristic separable. are What Do Inseparable Polynomials Look Like?
When char , inseparable polynomials are precisely the polynomials . After all, if is inseparable (and of the form for some therefore irreducible by definition), then , and this can happen only if the exponents of each term in are multiples of the characteristic . Hence, must have the form . But we can say more. Corollary 1.6.4 Let char is inseparable if and only if . An irreducible polynomial has the form over where and is a nonconstant polynomial. In this case, the integer can be chosen so that is separable, in which case every root of has multiplicity . In this case, the number is called the radical exponent of . Proof. As we mentioned, if is inseparable then , which implies that for all , which in turn implies that for all such that . Hence, . If has no multiple roots, we are done. If not, then we may repeat the argument with the irreducible polynomial , eventually obtaining the , where is separable. equation For the converse, suppose that in which both and split. Thus, for some . Let be a field for and so Polynomials 35 Since and so splits in , there exist roots . Hence, for each of the factors , This shows that is inseparable. Finally, if inseparable, then the 's above are distinct and so are the of has multiplicity . , where is 's. Hence, each root We can now prove that all irreducible polynomials over a finite field are separable. Corollary 1.6.5 All irreducible polynomials over a finite field are separable. Proof. First, we show that a finite field of characteristic has elements, for some . To see this, note that is an extension of its prime subfield and if the dimension of as a vector space over is , then has elements. It follows that the multiplicative group of nonzero elements of has order and so for all . In particular, any element of is a th power of some other element of . Thus, if is not separable, then . Hence is not irreducible. The next example shows that inseparable polynomials do exist. Example 1.6.1 Let be a field of characteristic and consider the field of 2 2 all rational functions in the variable . The polynomial is 2 2 irreducible over , since it has no linear factors over . However, in 2 we have and so is a double root of . 1.7 Testing for Irreducibility
We next discuss some methods for testing a polynomial for irreducibility. Note first that is irreducible if and only if is irreducible, for . This is often a useful device in identifying irreducibility. 36 Field Theory Localization
Sometimes it is possible to identify irreducibility by changing the base ring. In particular, suppose that and are rings and is a ring homomorphism. If a polynomial is degreewise reducible, then where deg deg and deg deg . Applying gives and since the degree cannot increase, if deg deg , then we can conclude that is degreewise reducible over . Hence, if is degreewise irreducible, then so is . This situation is a bit too general, and we take to be a field. Theorem 1.7.1 Let be a ring and let be a field. Let be a ring homomorphism. A polynomial is degreewise irreducible (not the product of two polynomials of smaller degree) over if 1) deg deg 2) is irreducible over . The following special case is sometimes called localization. Recall that if is a ring and is a prime, then the canonical projection map is defined by . This map is a surjective ring homomorphism. Corollary 1.7.2 (Localization) Let be a principal ideal domain and let be a polynomial over is irreducible over . Let , then be a prime that does not divide . If is degreewise irreducible over . Example 1.7.1 Let . Since has degree , it is reducible if and only if it has an integer root. We could simply start checking integers, but localization saves a lot of time. By localizing to , we get , and we need only check for a root in . Since none of these is a root, and therefore , is degreewise irreducible. But since is primitive, it is just plain irreducible. It is interesting to point out that there are polynomials for which is reducible for all primes , and yet is irreducible over . Thus, the method of localization cannot be used to prove that a polynomial is reducible. Example 1.7.2 Let is reducible for all primes polynomials , for , then . We claim that is one of the following . If Polynomials 37 or , each of which is reducible modulo . Now assume that , let satisfy , in which case written in any of the following ways . In the field , which can be Each of these has the potential of being the difference of two squares, which is reducible. In fact, this will happen if any of and is a square modulo . Since the multiplicative group of nonzero elements of is cyclic (a fact about finite fields that we will prove later), we can write . Note that the group homomorphism has kernel and so exactly half of the elements of are squares, and these are the even powers of . So, if and are nonsquares, that is, odd powers of , then their product is a square, and therefore so is Now, we can choose and so that is irreducible over . modulo . is irreducible over . For example, Eisenstein's Criterion
The following is the most famous criterion for irreducibility. Theorem 1.7.3 (Eisenstein's criterion) Let be an integral domain and let . If there exists a prime satisfying for , ,
2 then is degreewise irreducible. In particular, if is primitive, then it is irreducible. Proof. Let be the canonical projection map. Suppose that where deg deg and deg deg . Since for all , it follows that Since , this implies that and are monomials of positive 38 Field Theory
degree (since is an integral domain). In particular, the constant terms and are in , that is, and and therefore , which is a contradiction. Hence, is degreewise irreducible. Eisenstein's criterion can be useful as a theoretical tool. Corollary 1.7.4 Let be an integral domain that contains at least one prime. For every positive integer , there is an irreducible polynomial of degree over . Proof. According to Eisenstein's criterion, the primitive polynomial is irreducible, where is a prime. Exercises
1. 2. Prove that if is an integral domain, then so is (Chinese Remainder Theorem) Let relatively prime polynomials over a field . Let polynomials over . Prove that the system of congruences mod mod 3. has a unique solution modulo the polynomial . Let be fields with . Prove that if is a factorization of polynomials over , where two of the three polynomials have coefficients in , then the third also has coefficients in . Let be a unique factorization domain. Prove that for any and . Prove that if then the ring is not a principal ideal domain. Verify the division algorithm (Theorem 1.3.1) for commutative rings with identity. Hint: try induction on deg . Let . Prove that there exist polynomials , with deg deg and deg deg for which . be pairwise be 4. 5. 6. 7. if and only if and are not relatively prime. Let be the multiplicative group of all nonzero elements of a field . We have seen that if is a finite subgroup of , then is cyclic. Prove that if is an infinite field then no infinite subgroup of is cyclic. 9. Prove Theorem 1.5.1. 10. Show that the following are irreducible over . a) b) c) 8. Polynomials 39 d) 11. For 12. 13. 14. 15. 2 prime show that is irreducible over . Hint: apply Eisenstein to the polynomial . 2 Prove that for prime, is irreducible over . If is an infinite integral domain and is a polynomial in several variables over , show that is zero as a function if and only if it is zero as a polynomial. Let be a prime. Show that the number of monic irreducible polynomials of degree over is . There is a simple (but not necessarily practical) algorithm for factoring any polynomial over , due to Kronecker. In view of Theorem 1.2.3, it suffices to consider polynomials with integer coefficients. A polynomial of degree is completely determined by specifying of its values. This follows from the Lagrange Interpolation Formula Let be a polynomial of degree over . If has a nonconstant factor of degree at most , what can you say about the values for ? Construct an algorithm for factoring into irreducible factors. Use this method to find a linear factor of the polynomial over . 16. Prove that if , where each rational expression is in lowest common terms (no common nonconstant factors in the numerator and denominator) then and . 17. Let be a polynomial over with multiple roots. Show that there is a polynomial over whose distinct roots are the same as the distinct roots of , but that occur in only as simple roots. Reciprocal Polynomials
If is a polynomial of degree , we define the reciprocal polynomial by . Thus, if then If a polynomial satisfies , we say that is selfreciprocal. 18. Show that is a root of if and only if is a root of . 19. Show that the reciprocal of an irreducible polynomial with nonzero constant term is also irreducible. 40 Field Theory . Prove that if is a prime for which for , then is irreducible. 21. Show that if a polynomial is selfreciprocal and irreducible, then deg must be even. Hint: check the value of . 22. Suppose that , where and are irreducible, and is selfreciprocal. Show that either a) and with , or b) and for some . 20. Let Chapter 2 Field Extensions In this chapter, we will describe several types of field extensions and study their basic properties. 2.1 The Lattice of Subfields of a Field
If is an extension field of , then can be viewed as a vector space over . The dimension of over is denoted by and called the degree of over . A sequence of fields of fields, and we write for which is referred to as a tower The fact that dimension is multiplicative over towers is fundamental. Theorem 2.1.1 Let . Then Moreover, if is a basis for over a basis for over , then the set of products basis for over . Proof. For the independence of , suppose that
, , and is is a . Then , and the independence of over implies that for all , and the , independence of over implies that , for all and . Hence, is linearly independent. Next, if then there exist such that . Since each is a linear combination of the 's, it follows that is a linear combination of the products . Hence spans over . 42 Field Theory The Composite of Fields
If and are subfields of a field , then the intersection is clearly a field. The composite of and is defined to be the smallest subfield of containing both and . The composite is also equal to the intersection of all subfields of containing and . More generally, the composite of a family of fields, all of which are contained in a single field , is the smallest subfield of containing all members of the family. Note that the composite of fields is defined only when the fields are all contained in one larger field. Whenever we form a composite, it is with the tacit understanding that the relevant fields are so contained. A monomial over a family of fields with product of a finite number of elements from the union . is simply a The set of all finite sums of monomials over is the smallest subring of containing each field and the set of all quotients of elements of (the quotient field of ) is the composite . Thus, each element of involves only a finite number of elements from the union and is therefore contained in a composite of a finite number of fields from the family . The collection of all subfields of a field forms a complete lattice (under set inclusion), with meet being intersection and join being composite. The bottom element in is the prime subfield of (see Chapter 0) and the top element is itself. 2.2 Types of Field Extensions
Field extensions can be classified into several types, as shown in Figure 2.2.1. The goal of this chapter is to explore the properties of these various types of extensions. Field Extensions 43 Algebraic Transcendental Finitely generated algebraic (= finite) Finitely generated transcendental Simple algebraic Simple transcendental Base field F Figure 2.2.1 It is worth noting that some types of extensions are defined in terms of the individual elements in the extension, whereas others are more "global" in nature. For instance, an extension is algebraic if each element is algebraic over . Other characterizations involve properties of the field as a whole. For instance, is normal if is the splitting field of a family of polynomials over . Let us begin with the basic definitions (which will be repeated as we discuss each type of extension in detail). Recall that if , then an element is said to be algebraic over if is a root of some nonzero polynomial over . An element that is not algebraic over is said to be transcendental over . If and if is a subset of , the smallest subfield of containing both and is denoted by . When is a finite set, it is customary to write for . Definition Let . Then 1) is algebraic over if every element is algebraic over . Otherwise, is transcendental over . 2) is finitely generated over if , where is a finite set. 3) is a simple extension of if , for some . In this case, is called a primitive element of . 4) is a finite extension of if is finite. To save words, it is customary to say that the extension is algebraic, transcendental, finitely generated, finite or simple, as the case may be, if has this property as an extension of . The reader may have encountered a different meaning of the term primitive in connection with elements of a finite field. We will discuss this alternative meaning when we discuss finite fields later in the book. 44 Field Theory Note that a transcendental extension may have algebraic elements not in the base field. For example, the transcendental extension has many algebraic elements, such as . In later chapters, we will study two other extremely important classes of extensions: the separable and the normal extensions. Briefly, an algebraic element is separable over if its minimal polynomial is separable and an extension is separable if every element of is separable over . When char or when is a finite field, all algebraic extensions are separable, but such is not the case with more unusual fields. As mentioned earlier, an extension of is normal if it is the splitting field of a family of polynomials. An extension that is both separable and normal is called a Galois extension. Distinguished Extensions
We will have much to say about towers of fields of the form . Let us refer to such a tower as a 2tower, where is the intermediate field, is the lower step, is the upper step and is the full extension. Following Lang, we will say that a class of field extensions is distinguished provided that it has the following properties 1) The Tower Property For any 2tower , the full extension is in and lower steps are in . In symbols, and 2) The Lifting Property The class is closed under lifting by an arbitrary field, that is, and provided, of course, that by . Note that if is defined. The tower is the lifting of if and only if the upper is distinguished, then it also has the following property: 3) Closure under finite composites If is defined, then and This follows from the fact that can be decomposed into Field Extensions 45 and the first step is in , the second step is in by , and so the full extension is in . Figure 2.2.2 illustrates these properties. since it is the lifting of E EK EK K E F
Composite K F
Tower E F
Lifting K Figure 2.2.2 Consider a tower of field extensions We say that the tower is in , or has property , if all extensions of the form , where , are in . To illustrate the terminology, an algebraic tower is a tower in which each extension , where , is algebraic. If a class tower has the tower property, then the following are equivalent for a finite : 1) is in 2) The full extension 3) Each step If a class is in is in . of extensions has the property that for any family of fields (provided, as always, that the composite is defined), we say that is closed under arbitrary composites. This property does not follow from closure under finite composites. Here is a list of the common types of extensions and their distinguishedness. We will verify these statements in due course. Distinguished Algebraic extensions Finite extensions Finitely generated extensions Separable extensions 46 Field Theory Not Distinguished Simple extensions (lifting property holds, upper and lower steps simple) Transcendental extensions Normal extensions (lifting property holds, upper step normal) 2.3 Finitely Generated Extensions
If and if is denoted by write is a subset of . When for . , the smallest subfield of containing and is a finite set, it is customary to Definition Any field of the form generated over and the extension Any extension of the form called a primitive element in . is said to be finitely is said to be finitely generated. is called a simple extension and is The reader may have encountered a different meaning of the term primitive in connection with elements of a finite field. We will discuss this alternative meaning when we discuss finite fields later in the book. Note that for , and so a finitely generated extension into a tower of simple extensions can be decomposed It is evident that 's: consists of all quotients of polynomials in the The class of finite extensions is our first example of a distinguished class. Theorem 2.3.1 The class of all finitely generated extensions is distinguished. Proof. For the tower property, if is a 2tower in which each step is finitely generated, that is, if and are finite sets, then since , the full extension is finitely generated by over . Also, if , where is finite, then since , the upper step is , which is finitely generated by . However, the proof that the lower step is finitely generated is a bit testy and we must postpone it until we have discussed transcendental extensions in the next chapter. Field Extensions 47 For the lifting property, if defined, then , where is finite and if , with and so the composite is finitely generated over by . 2.4 Simple Extensions
Let us take a closer look at simple extensions . Simple Extensions Are Not Distinguished
The class of simple extensions has all of the properties required of distinguished extensions except that the lower and upper steps being simple does not imply that the full extension is simple. That is, if each step in a 2tower is simple this does not imply that the full extension is simple. Example 2.4.1 Let tower and be independent variables and let be a prime. In the each step is simple but the full extension is not. We leave proof of this as a (nontrivial) exercise. On the other hand, if the full extension is simple , then the upper step is , which is simple. Also, the lower step is simple, but the nontrivial proof requires us to consider the algebraic and transcendental cases separately, which we will do at the appropriate time. As to lifting, if is simple and , then the lifting is which is simple. Thus, the lifting property holds. , Simple Algebraic Extensions
Suppose that is a simple extension, where is algebraic over . We have seen that the minimal polynomial of over is the min unique monic polynomial of smallest degree satisfied by . Also, is irreducible. Now, is the field of all rational expressions in but we can improve upon this characterization considerably. Since , it follows that and the irreducibility of implies that 48 Field Theory polynomials . Hence, and and for which are relatively prime and there exist Evaluating at gives and so the inverse of is the polynomial . It follows that Moreover, if deg deg , then where or deg deg . Hence, Thus, deg deg In words, is the set of all polynomials in over of degree less than the degree of the minimal polynomial of , where multiplication is performed modulo . The map defined by mod is easily seen to be a surjective ring homomorphism. In fact, it is the composition of two surjective ring homomorphisms: the first is projection modulo and the second is evaluation at . The kernel of is the ideal ker generated by mod , since It follows that This has a couple of important consequences. First, if we restrict attention to polynomials of degree less than deg , then can be treated as an "independent" variable. Also, if are conjugate (have the same minimal Field Extensions 49 polynomial) over , then the substitution map defined by is an isomorphism from Let us summarize. Theorem 2.4.1 Let 1) Then to . and let be algebraic over deg deg . . where multiplication is performed modulo 2) Moreover, 3) The extension is finite and deg In fact, the set 4) If the elements is a vector space basis for are conjugate over then over . . We have seen that a simple extension , where is algebraic, is finite. Conversely, if is finite and simple, then for any , the sequence is linearly dependent and so is algebraic. Hence, all elements of are algebraic and so is an algebraic extension. Theorem 2.4.2 The following are equivalent for a simple extension 1) is algebraic 2) is algebraic 3) is finite. In this case, deg min . Characterizing Simple Algebraic Extensions Simple algebraic extensions can be characterized in terms of the number of intermediate fields. Theorem 2.4.3 Let be finitely generated over by algebraic elements over 1) Then for some algebraic element if and only if there is only a finite number of intermediate fields between and . 50 Field Theory 2) In this case, if is an infinite field, then where has the form for . Proof. Suppose first that for some algebraic element . For each intermediate field , the minimal polynomial min is also a polynomial over and is satisfied by . Hence, min min . But min has only a finite number of monic factors. Therefore, this part of the proof will be complete if we show that there is only one intermediate field with minimal polynomial min . Suppose that and have the property that min Then the coefficients of is also irreducible over min But and so deg which implies that and so . Similarly, and so . This shows that is uniquely determined by the polynomial min and so there are only finitely many intermediate fields . For the converse, if is a finite field, then so is , since it is finitedimensional over and so the multiplicative group of nonzero elements of is cyclic. If generates this group, then is simple. Now suppose that is an infinite field and there are only finitely many intermediate fields between and . Consider the intermediate fields , for all . By hypothesis, for some . Hence, , implying that lie in and so min min . Since min is irreducible over , it and Hence, . Hence, . The reverse inclusion is evident and so Field Extensions 51 We can repeat this process to eventually arrive at a primitive element of the desired form. In view of the previous theorem, it is clear that if , where is algebraic, then the lower step is also simple. (Note that is a finite extension and therefore finitely generated by the elements of a basis for over , whose elements are algebraic over .) Simple Transcendental Extensions
If is transcendental over : , then is the field of all rational expressions in The fact that is transcendental implies that there are no algebraic dependencies in these rational expressions and is, in fact, isomorphic to the field of rational functions in a single variable. Theorem 2.4.4 Let and let be transcendental over . Then is isomorphic to the field of all rational functions in a single variable . Proof. The evaluation homomorphism defined by is easily seen to be an isomorphism. To see that implies , which implies that would be algebraic. is injective, note that , since otherwise Simple transcendental extensions fail rather misreably to be distinguished. For example, the lifting of the transcendental extension by is , which is algebraic. Also, in the tower , the upper step is algebraic. Let be the field of rational functions in two independent variables. Then each step in the 2tower is simple, but the extension is not simple. The proof is left as an exercise. (Intuitively speaking, we cannot expect a single rational function in and to be able to express both and individually.) On the other hand, the lower step of a transcendental extension is simple and transcendental (provided that . This result is known as Luroth's theorem and will be proved in the next chapter. 52 Field Theory Thus, simple transcendental extensions fail to be distinguished on every count except that the lower step in a simple transcendental extension is simple and transcendental. More on Simple Transcendental Extensions The fact that the upper step in the tower is algebraic is not an isolated case. Suppose that is transcendental. Then any is a nonconstant rational function in where we can assume that and are relatively prime. It turns out that carries with it the full "transcendental nature" of the extension . To be more precise, consider the polynomial Then is a root of upper step in the tower and so is algebraic over . In other words, the is algebraic and finitely generated (by ) and therefore finite, by Theorem 2.4.2. As to the lower step, if it were also algebraic, it would be finite and so by the multiplicativity of degree, would be finite and therefore algebraic. Since this is not the case, we deduce that is transcendental, which means that does not satisfy any nonzero polynomial over . We can now show that is irreducible over . Since is transcendental over , we have , where is an independent variable. It follows that and so it is sufficient to show that the polynomial is irreducible over . However, this follows from the fact that is irreducible as a polynomial over the ring , that is, as a polynomial in . To see this, note that any factorization in has the form where and so . and are over . But and must be a unit in , which implies that are relatively prime is irreducible over Field Extensions 53 Hence, is irreducible over deg and max deg , deg . Let Theorem 2.4.5 1) Consider the extension , where is transcendental over be any element of in the tower , where and are relatively prime. Then the lower step is transcendental (and so is transcendental over upper step is algebraic, with max deg , deg ) and the 2) If is transcendental over , then is algebraic over any intermediate field other than itself. Proof. Part 1) has already been proved. As to part 2), if where , then let . In the tower , we know that is algebraic and simple and thus finite. It follows that is also finite, hence algebraic. We should note that this theorem does not hold for nonsimple extensions. Specifically, just because an extension is generated by transcendental elements does not mean that all of the elements of are transcendental. For example, the extension , where is transcendental over , is generated by transcendental elements and , but some elements of are algebraic over . We will have more to say about this in Chapter 3. 2.5 Finite Extensions
If and is finite. is finite, we say that is a finite extension of or that Theorem 2.5.1 An extension is finite if and only if it is finitely generated by algebraic elements. Proof. If is finite and if is a basis for over , then is finitely generated over . Moreover, for each , the sequence over powers is linearly dependent over , and so is algebraic over . Thus, is algebraic. For the converse, assume that over . Each step in the tower , , where each is algebraic 54 Field Theory is simple and algebraic, hence finite by Theorem 2.4.2. It follows that over . Suppose that over and consider the tower is finite is finitely generated by algebraic elements Our results on simple algebraic extensions show that any element of is a polynomial in over . Further, any element of is a polynomial in over , and hence a polynomial in the two variables and . Continuing in this way, we conclude that is the set of all polynomials over in . Theorem 2.5.2 The class of finite extensions is distinguished. Moreover, if is a finite basis for over and if , then spans over , in particular, Proof. The multiplicativity of degree shows that the tower property holds. As to lifting, let be finite, with basis and let . Then , , where each is algebraic over and so also over . Since , is finitely generated by elements algebraic over , it is a finite extension of . For the statement concerning degree, let be a basis for over . If , then the lifting is and each is algebraic over . It follows that is the set of polynomials over in . However, any monomial in the 's is a linear combination (over of and so is the set of linear combinations of over . In other words, spans over . We will see much later in the book that if separable, then actually divides is finite, and also normal and . Note that if is a splitting field for , then is generated by the set of distinct roots of . Thus is finitely generated by algebraic elements and so is a finite extension of , of degree at most , where deg . 2.6 Algebraic Extensions
We now come to algebraic extensions. Field Extensions 55 Definition An extension of algebraic over . Otherwise, is algebraic over if every element is transcendental over . is Theorem 2.6.1 A finite extension is algebraic. Proof. As we have said before, if is finite and 2 of powers is linearly dependent over nontrivial polynomial in must equal , implying that , then the sequence and therefore some is algebraic over . Corollary 2.6.2 The following are equivalent for an extension 1) is finite 2) is finitely generated by algebraic elements 3) is algebraic and finitely generated. Theorem 2.6.3 Let . The set of all elements of that are algebraic over is a field, called the algebraic closure of in . Proof. Let . The field is finitely generated over by algebraic elements and so is algebraic over , that is, . This implies that , and all lie in , and so is a subfield of . Theorem 2.6.4 The class of algebraic extensions is distinguished. It is also closed under the taking of arbitrary composites. Proof. For the tower property, let . If the full extension is algebraic then so is the lower step . Also, since any polynomial over is a polynomial over , the upper step is also algebraic. Conversely, suppose that and are algebraic and let have minimal polynomial over . Consider the tower of fields Since over is algebraic over and each , being in , is algebraic , we deduce that each step in the tower is finite and so is finite. Hence, is algebraic over . For the lifting property, let be algebraic and let . Let , where is the algebraic closure of in . Then since each is algebraic over it is a fortiori algebraic over and so . Clearly, and so . It follows that is algebraic over . Finally, if is a family of fields, each algebraic over , then so is , since an element of is also an element of a composite of only a finite number of members of the family. The algebraic closure of the rational numbers in the complex numbers is called the field of algebraic numbers. We saw in the previous chapter that there is an irreducible polynomial of every positive degree . 56 Field Theory Hence, is an infinite algebraic extension of , showing that the converse of Theorem 2.6.1 does not hold: algebraic extensions need not be finite. Note that if is an algebraic number, it also satisfies a polynomial over the integers. Thus, the algebraic numbers can be defined as the set of complex roots of polynomials over the integers. The subfield of all complex roots of monic polynomials over the integers is called the field of algebraic integers. We note finally that if is algebraic and if for some then each element of is a polynomial in finitely many elements from . This follows from the fact that each is a rational function in finitely many elements of and so there exists a finite subset such that . Hence, our discussion related to finitely generated algebraic extensions applies here. 2.7 Algebraic Closures
Definition A field is said to be algebraically closed if any nonconstant polynomial with coefficients in splits in . Note that an algebraically closed field cannot have a nontrivial algebraic extension , since any is algebraic over and its minimal polynomial over must split over , whence . Theorem 2.7.1 Let be a field. Then there is an extension of that is algebraically closed. Proof. The following proof is due to Emil Artin. The first step is to construct an extension field of , with the property that all nonconstant polynomials in have a root in . To this end, for each nonconstant polynomial , let be an independent variable and consider the ring of all polynomials in the variables over the field . Let be the ideal generated by the polynomials . We contend that is not the entire ring . For if it were, then there would exist polynomials and such that This is an algebraic expression over in a finite number of independent variables. But there is an extension field of in which each of the polynomials has a root, say . Setting and setting any other variables appearing in the equation above equal to gives . This contradiction implies that . Since , there exists a maximal ideal such that . Then is a field in which each polynomial has a root, namely . (We may think of as an extension of by identifying with .) Field Extensions 57 Using the same technique, we may define a tower of field extensions such that each nonconstant polynomial has a root in . The union is an extension field of . Moreover, any polynomial has all of its coefficients in for some and so has a root in , hence in . It follows that every polynomial splits over . Hence is algebraically closed. Definition Let . Then is an algebraic closure of if is algebraic and is algebraically closed. We will denote an algebraic closure of a field by . We can now easily establish the existence of algebraic closures. Theorem 2.7.2 Let where is algebraically closed. Let where is the algebraic closure of in . Then is the only algebraic closure of that is contained in . Thus, any field has an algebraic closure. Proof. We have already seen that is an algebraic extension of . By hypothesis, any splits in and so all of its roots lie in . Since these roots are algebraic over , they are also algebraic over and thus lie in . Hence splits in and so is algebraically closed. As to uniqueness, if with an algebraic closure of , then since is algebraic, we have . But if the inclusion is proper, then there is an . It follows that min does not split over , a contradiction to the fact that is algebraically closed. Hence, . The final statement of the theorem follows from Theorem 2.7.1. We will show a bit later in the chapter that all algebraic closures of a field are isomorphic, which is one reason why the notation is (at least partially) justified. Here is a characterization of algebraic closures. Theorem 2.7.3 Let . The following are equivalent. 1) is an algebraic closure of . 2) is a maximal algebraic extension of , that is, is algebraic and if is algebraic then . 3) is a minimal algebraically closed extension of , that is, if where is algebraically closed, then . 4) is algebraic and every nonconstant polynomial over splits over . 58 Field Theory Proof. To see that 1) implies 2), suppose that is an algebraic closure of is algebraic. Hence, any is algebraic over . But min splits over and so contains a full set of roots of min . Hence, which shows that . Thus, is a maximal algebraic extension of . and , Conversely, let be a maximal algebraic extension of and let . Let be the splitting field for over . Thus, is an algebraic tower, since is generated over by the finite set of roots of . Hence, the maximality of implies that , and so splits in , which says that is algebraically closed and therefore an algebraic closure of . To see that 1) implies 3), suppose that where is algebraically closed. Since is algebraic, it follows that . Conversely, suppose that is a minimal algebraically closed extension of . Let be the algebraic closure of in . Thus, , with algebraic. If is not algebraically closed, then there is a polynomial over that does not split over . But is also a polynomial over and therefore splits over . Hence, each of its roots in is algebraic over and therefore also over , and so lies in , which is a contradiction. Hence, is algebraically closed and so the minimality of implies that , whence is an algebraic closure of . Finally, it is clear that 1) implies 4). If 4) holds, then is algebraic and if is algebraic, then let have minimal polynomial over . This polynomial splits over and so , which implies that , whence is a maximal algebraic extension of and so 2) holds. 2.8 Embeddings and Their Extensions
Homomorphisms between fields play a key role in the theory. Since a field has no ideals other than and , it follows that any nonzero ring homomorphism from into a ring must be a monomorphism, that is, an embedding of into . A bit of notation: Let be a function. 1) The restriction of to is denoted by . 2) The image of under is denoted by or by . 3) The symbol denotes an embedding. Thus, signifies that is an embedding of into . 4) If and if is an embedding, the polynomial is denoted by or . Definition Let be an embedding of into and let . Referring to Figure 2.8.1, an embedding for which is called an extension of to . An embedding of that extends the identity map is called an embedding over , or an embedding. Field Extensions 59 L (E) E F
Figure 2.8.1 (F) The set of all embeddings of into is denoted by hom embeddings of into that extend is denoted by hom all embeddings over is denoted by hom . . The set of all and the set of Embeddings play a central role in Galois theory, and it is important to know when a given embedding can be extended to a larger field , and how many such embeddings are possible. We will discuss the former issue here, and the latter issue in the next chapter. The Properties of Embeddings
Embeddings preserve many properties. For example, an embedding maps roots to roots and preserves composites. Lemma 2.8.1 1) (Embeddings preserve factorizations and roots) If , then if and only if Also, is a root of if and only if is a root of 2) (Embeddings preserve the lattice structure) If is a family of subfields of then and 3) (Embeddings preserve adjoining) If then and if and and . . and if 4) (Embeddings preserve being algebraic) Let and let be algebraic. If is an extension of , then is algebraic. 5) (Embeddings preserve algebraic closures) Let and let be an algebraic closure of . If is an extension of , then is an algebraic closure of . Proof. We leave the proof of parts 1), 4) and 5) to the reader. For part 2), since is injective, it preserves intersections. But for all 60 Field Theory and so for all for all For part 3), contains and and so it contains the smallest field containing these two sets, that is . On the other hand, if is a field for which and , then and , whence , and so . In other words, is contained in any field containing and , including the composite . Even though the next result has a simple proof, the result is of major importance. If is algebraic and over , then since permutes the roots of any polynomial over and since every element of is a root of a polynomial over , it follows that every element of is the image of some element of , that is, the embedding must be surjective, and hence an automorphism. Theorem 2.8.2 If is algebraic and automorphism of . In symbols, hom Aut over , then is an Proof. Let and let be the set of roots of the minimal polynomial min that lie in . Then is a permutation on and so there is a for which . Hence, is surjective and thus an automorphism of . Extensions in the Simple Case
Consider the case of a simple algebraic extension. Suppose that , where is algebraically closed. Let be algebraic over . We can easily extend to , using the minimal polynomial of over . The key point is that any extension of on and this value must be a root of we must have is completely determined by its value min , where . In fact, for any isomorphism . Moreover, it is easy to see that this condition defines an over . and let be algebraic over , with minimal , where is algebraically closed. in , then can be extended to an embedding for which . Theorem 2.8.3 Let polynomial . Let 1) If is a root of over Field Extensions 61 2) Any extension of to must have the form , as described in part 1). 3) The number of extensions of to is equal to the number of distinct roots of min in . The previous theorem shows that the cardinality of hom depends only on through its minimal polynomial, and furthermore, that it does not depend on either or ! We will explore this issue further in the next chapter. Extensions in the Algebraic Case
The simple case, together with Zorn's lemma, is just what we need to prove that if , with algebraically closed and if is algebraic, then there is at least one extension of to . Theorem 2.8.4 Let be algebraic. 1) Any embedding , where is algebraically closed, can be extended to an embedding . 2) Moreover, if and min and is a root of , then we can choose so that . (See Figure 2.8.2.) Proof. Let be the set of all embeddings hom for which and where . Since can be extended to an embedding of into in such a way that , it follows that is not empty. is a partially ordered set under the order defined by saying that if and is an extension of . If is a chain in , the map defined by the condition , is an upper bound for in . Zorn's lemma implies the existence of a maximal extension . We contend that , for if not, there is an element . But is algebraic over and so we may extend to , contradicting the maximality of . L The set E (E) F( ) F( ) F
Figure 2.8.2 (F) As a corollary, we can establish the essential uniqueness of algebraic closures. Corollary 2.8.5 Any two algebraic closures of a field are isomorphic. 62 Field Theory Proof. Let and be algebraic closures of . The identity map can be extended to an embedding . Since is algebraically closed so is . But is an algebraic extension of and so . Hence, is an isomorphism. Independence of Embeddings
Next, we come to a very useful result on independence of embeddings. We choose a somewhat more general setting, however. A monoid is a nonempty set with an associative binary operation and an identity element. If and are monoids, a homomorphism of into is a map such that and . Definition Let be a monoid and let be a field. A homomorphism , where is the multiplicative group of all nonzero elements of is called a character of in . Note that an embedding . of fields is a character, when restricted to Theorem 2.8.6 ( . Artin) Any set of distinct characters of independent over . Proof. Suppose to the contrary that in is linearly for and , not all . Look among all such nontrivial linear combinations of the 's for one with the fewest number of nonzero coefficients and, by relabeling if necessary, assume that these coefficients are . Thus, (2.8.1) for all and this is the "shortest" such nontrivial equation (hence for all ). Note that since , we have for all . Hence, . Let us find a shorter relation. Since Multiplying by gives is a character, for all . On the other hand, replacing by in (2.8.1) gives, Subtracting the two equations cancels the first term, and we get Field Extensions 63 Now, since , there is an for which and we have a shorter nontrivial relation of the form (2.8.1). This contradiction proves the theorem. Corollary 2.8.7 (Dedekind independence theorem) Let and be fields. Any set of distinct embeddings of into is linearly independent over . 2.9 Splitting Fields and Normal Extensions
Let us repeat a definition from Chapter 1. Definition Let be a family of polynomials in . A splitting field for over is an extension field of with the property that each splits in and that is generated by the set of all roots of the polynomials in . The next theorem says that splitting fields not only exist, but are essentially unique. Theorem 2.9.1 (Existence and uniqueness of splitting fields) Let be a family of polynomials over . 1) In any algebraic closure of , there is a unique splitting field for . 2) If and are algebraic, where is the splitting field for in and is the splitting field for in then any embedding over maps onto . 3) Any two splitting fields for are isomorphic over . Proof. For part 1), if is a family of polynomials over , then every member of splits in and so contains the field generated over by the roots in of the polynomials in , that is, contains a splitting field for . It is clear that this splitting field is unique in , because any splitting field in must be generated, in , by the roots of all polynomials in . For part 2), if so is the family of roots of contained in then and But is precisely the set of roots of field for in , that is, from part 2). in and so is the splitting . Part 3) follows immediately Invariance and Normal Extensions
Speaking very generally, if is any function on a set and if has the property that , then is said to be invariant under , or invariant. This notion occurs in many contexts, including the present one, although the term "invariant" is seldom used in the present context. 64 Field Theory Suppose that is algebraic and that is an embedding over of into the algebraic closure . Then is invariant if . However, since is algebraic, any embedding of into itself is an automorphism of and so is invariant if and only if , that is, if and only if is an automorphism of . Suppose that is invariant for all embeddings over . Then it is not hard to see that any irreducible polynomial over that has one root in must split over . For if is also a root of in , then there is an embedding hom for which . Hence, the invariance of implies that . Put another way, we can say that is the splitting field for the family MinPoly Thus, for min 2) 3), where algebraic, we have shown that 1) 1) is invariant for all embeddings over 2) If an irreducible polynomial over has one root in , then it splits over 3) is a splitting field, specifically for the family MinPoly . . On the other hand, suppose that is a splitting field of a family of polynomials over . Thus, , where is the set of roots of the polynomials in . But any embedding over sends roots to roots and so sends to itself. Hence, Since is an embedding of into itself over and is algebraic, it follows that is an automorphism of . Thus, 1)3) are equivalent. Theorem 2.9.2 Let , where is an algebraic closure of . The following are equivalent. 1) is a splitting field for a family of polynomials over . 2) is invariant under every embedding over . (It follows that every embedding of into over is an automorphism of .) 3) Every irreducible polynomial over that has one root in splits in . Definition An algebraic extension that satisfies any (and hence all) of the conditions in the previous theorem is said to be a normal extension and we write . We also say that is normal over . Corollary 2.9.3 If is a finite normal extension, then field of a finite family of irreducible polynomials. is the splitting Field Extensions 65 Proof. Let . Since , each minimal polynomial min splits in . Clearly, is generated by the roots of the finite family min and so is the splitting field of . Note that the extension splits in . is normal, since any nonconstant Normal Extensions Are Not Distinguished
As it happens, the class of normal extensions is not distinguished, but it does enjoy some of the associated properties. Example 2.9.1 It is not hard to see that any extension of degree is normal. The extension is not normal since contains exactly two of the four roots of the irreducible polynomial . On the other hand, has each step of degree and therefore each step is normal. Here is what we can say on the positive side. Theorem 2.9.4 1) (Full extension normal implies upper step normal) Let . If is normal then . 2) (Lifting of a normal extension is normal) If and then . 3) (Arbitrary composites and intersections of normal are normal) If is a family of fields, and then and . Proof. Part 1) follows from the fact that a splitting field for a family of polynomials over is also a splitting field for the same family of polynomials over . For part 2), let be a splitting field for a family of polynomials over and let be the set of roots in of all polynomials in . Then . Hence, , which shows that is a splitting field for the family , thought of as a family of polynomials over . Hence, . For part 3), let each and so over . Then , whence is an embedding when restricted to and so is an automorphism of . Similarly, if over then 66 Field Theory Normal Closures
If is not normal, then there is a smallest extension of (in a given algebraic closure ) for which is normal over . Perhaps the simplest way to see this is to observe that is normal and the intersection of normal extensions is normal, so and Definition Let . The normal closure of smallest intermediate field for which denoted by nc . Theorem 2.9.5 Let 1) The normal closure nc over in is the . The normal closure is be algebraic, with normal closure nc exists and is equal to and . 2) nc
hom 3) nc is the splitting field in MinPoly of the family min 4) If family , where MinPoly , then nc min is the splitting field in of the 5) If is finite, then nc is also finite. Proof. We prove only part 2), leaving the rest for the reader. Let with . Since is algebraic, any embedding hom can be extended to an embedding over . Since , is an automorphism of . It follows that and so . On the other hand, if , then , since if hom then runs over all elements of hom as does and so Hence, nc and . is the smallest normal extension of in , that is, Exercises
1. Prove that . Field Extensions 67 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. Prove that if is an algebraic extension of the real field and , then is isomorphic to the complex numbers . Prove that every finite field of characteristic is a simple extension of its prime subfield . Let . Suppose that is finite, where is a subset of . Is it true that for some finite subset of ? If and is algebraically closed, is necessarily an algebraic closure of ? Suppose that char and that . Let . Prove that if min has only one distinct root in , then and the multiplicity of is . What can be said if char ? Let be a quadratic extension, that is, an extension of degree . Show that has a basis over of the form where 2 . a) Find all automorphisms of . b) Is there an isomorphism over for which ? c) Is there an isomorphism over other than the identity? Show that the automorphism over that sends to is not continuous. Prove that if is algebraic and has only a finite number of intermediate fields, then is a finite extension. Let be an integral domain containing a field . Then is a vector space over . Show that if then must be a field. Find a counterexample when is a commutative ring with identity but not an integral domain. If is algebraic and is a ring such that , show that is a field. Is this true if is not algebraic? Let and be finite extensions and assume that is defined. Show that , with equality if and are relatively prime. Let and let . Show that is linear if and only if for all . Find an extension that is algebraic but not finite. The algebraic closure of in , that is, the set of all complex roots of polynomials with integer coefficients, is called the field of algebraic numbers. Prove that is algebraic and infinite by showing that if are distinct primes, then Hint: use induction on . 17. Prove that any extension of degree is normal. 18. Let be a finite Galois extension and let and over , respectively. Suppose that have degrees . 68 Field Theory Show that if is a conjugate of and is a conjugate of , then there is a such that and . Hence, the conjugates of are . b) Show that if the difference of two conjugates of is never equal to the difference of two conjugates of then . 19. Let be an infinite field and let be an algebraic extension. Show that . 20. Let where is an algebraic closure of and let Aut be the group of all automorphisms of fixing pointwise. Assume that all irreducible polynomials over are separable. Let fix for all a) 21. 22. 23. 24. 25. 26. 27. 28. be the fixed field of under . Evidently fix . Prove that fix . (For readers familiar with complex roots of unity) Let be a prime and let be a complex th root of unity. Show that min . What is the splitting field for over ? Let be a field of characteristic and let . Show that the following are equivalent: a) , b) , c) where . Let be a finite normal extension and let be irreducible. Suppose that the polynomials and are monic irreducible factors of over . Show that there exists a Aut for which . Show that an extension is algebraic if and only if any subalgebra of over is actually a subfield of . Let . Can all automorphisms of be extended to an automorphism of ? Suppose that and are fields and is an embedding. Construct an extension of that is isomorphic to . Let be algebraic. a) Finish the proof of Theorem 2.9.5. b) Show that any two normal closures and , where is an algebraic closure of are isomorphic. With reference to Example 2.4.1, let and be independent variables and let be a prime. Show that, in the tower each step is simple but the full extension is not. 29. Consider the field of rational functions in two (independent) variables. Show that the extension is not simple. Field Extensions 69 Constructions
The goal of the following series of exercises is to prove that certain constructions are not possible using straight edge and compass alone. In particular, not all angles can be trisected, a circle cannot be "squared" and a cube cannot be "doubled." The first step is to define the term constructible. Definition We assume the existence of two distinct points and in the plane and take the distance between these points to be one unit. A point, line or circle in the plane is said to be constructible if it can be obtained by a finite number of applications of the following rules: 1) and are constructible. 2) The line through any two constructible points is constructible. 3) The circle with center at one constructible point and passing through another constructible point is constructible. 4) The points of intersection of any two constructible lines or circles are constructible. 30. Show that if a line and point are constructible, then the line through perpendicular to is also constructible. 31. Show that if a line and point are constructible, then the line through parallel to is also constructible. 32. Taking the constructible line through and as the axis and the point as the origin, the axis is also constructible. Show that any point with integer coordinates is constructible. 33. Show that the perpendicular bisector of any line segment connecting two constructible points is constructible. Show that the circle through two constructible points and with center equal to the midpoint of and is constructible. 34. If , and are constructible points and is a constructible line through then a point can be constructed on such that the distance from to is the same as the distance from to . (Thus, given distances can be marked off on constructible lines.) Constructible Numbers Definition A real number is constructible if its absolute value is the distance between two constructible points. 35. Show that any integer is constructible. 36. Prove that a point is constructible if and only if its coordinates and are constructible real numbers. 37. Prove that the set of numbers that are constructible forms a subfield of the real numbers containing . Hint: to show that the product of two constructible numbers is constructible or that the inverse of a nonzero constructible number is constructible, use similar triangles. 70 Field Theory 38. Prove that if 0 is constructible, then so is . Hint: first show that a circle of diameter , with center on the axis and going through the origin is constructible. Mark off units along the axis and draw the perpendicular. The two previous exercises prove the following theorem. Theorem C1 If the elements of a field are constructible, and if then is constructible. Theorem C2 Let be a subfield of Then for some . Proof. Exercise. and let , be a quadratic extension. It follows from the two previous theorems that if is constructible and if is a quadratic extension, that is, , then is constructible. Any tower , where each extension has degree is a quadratic tower. Thus, if is a quadratic tower, then every element of is constructible. The converse of this statement also happens to be true. Theorem C3 (Constructible numbers) The set of constructible real numbers is the set of all numbers that lie in some quadratic tower with base . In particular, the degree of a constructible number over a power of . Proof. Exercise. must be Constructible Angles Now consider what it means to say that an angle of is constructible. Informally, we will take this to mean that we may construct a line through the origin that makes an angle of with the axis. Formally, the angle (real number) is constructible if the real number cos is constructible. 39. Show that such a line making angle with the axis is constructible if and only if the real number cos is constructible. (This is an informal demonstration, since we have not formally defined angles.) 40. Show that a angle is constructible. 41. Show that a angle is not constructible. Hint: Verify the formula cos Let cos and show that cos3 is a root of cos Field Extensions 71 3 Show that is irreducible over and so . 42. Prove that every constructible real number is algebraic over . Assuming that is transcendental over , show that any circle with a constructible radius cannot be "squared," that is, a square cannot be constructed whose area is that of a unit circle. 43. Verify that it is impossible to "double" any cube whose side length is constructible, that is, it is impossible to construct an edge of a cube whose volume is twice that of a cube with side length . Chapter 3 Embeddings and Separability 3.1 Recap and a Useful Lemma
Let us recall a few facts about separable polynomials from Chapter 1. Definition An irreducible polynomial is separable if it has no multiple roots in any extension of . An irreducible polynomial that is not separable is inseparable. Definition If , then an algebraic element is separable if its minimal polynomial min is separable. Otherwise, it is inseparable. Also, the radical exponent of over is the radical exponent of min . Theorem 3.1.1 1) An irreducible polynomial is separable if and only if . 2) If is a field of characteristic , or a finite field, then all irreducible polynomials over are separable. 3) Let char . An irreducible polynomial over is inseparable if and only if has the form where and is a nonconstant polynomial. In this case, the integer can be chosen so that is separable and then every root of has multiplicity , where is called the radical exponent of . The radical exponent of can be characterized as the largest integer for which . 4) Let char . If has radical exponent then is separable over , and is the smallest power of for which is separable over . 74 Field Theory In Chapter 2, we considered the problem of extending the domain of an embedding of to a larger field that is algebraic over . Here is a brief summary of what we discussed. Theorem 3.1.2 1) (Simple extensions) Let and let be algebraic over , with minimal polynomial . Let , where is algebraically closed. a) If is a root of in , then can be extended to an embedding over for which . b) Any extension of to must have the form . c) The number of extensions of to is equal to the number of distinct roots of min in . 2) (Algebraic extensions) Let be algebraic. Any embedding , where is algebraically closed, can be extended to an embedding . Moreover, if , min and is a root of , then we can choose so that . A Useful Lemma
Before proceeding, we record a useful lemma. If denotes the set . is a field and then Lemma 3.1.3 Let be algebraic with char and let . 1) holds for some if and only if it holds for all . 2) holds for some if and only if it holds for all . Proof. For part 1), suppose that holds for some . Since it follows that . Now, since , we have for any and so [ Hence, For part 2), we observe that ] , for all . and so only if holds for some for all . if and only if , which holds if and Embeddings and Separability 75 3.2 The Number of Extensions: Separable Degree
According to Theorem 2.8.3, the number of extensions of an embedding to , where is algebraically closed, is equal to the number of distinct roots of min . Hence, as we remarked earlier, the size of does not depend on either or . The same is true for hom extensions of to any algebraic extension. Theorem 3.2.1 If is algebraic and , where is algebraically closed, then the cardinality of hom depends only on the extension and not on or . In other words, if , with algebraically closed, then hom hom as cardinal numbers. Proof. We refer the reader to Figure 3.2.1. Since for any hom , the image is contained in an algebraic closure of , we may assume that is an algebraic closure of , and similarly, that is an algebraic closure of . Since is an isomorphism and is algebraic, the map can be extended to an embedding of into Since is algebraic, so is its image under , which is , and since is algebraically closed, we have , implying that is an isomorphism. Now, if hom , then the map is an embedding of into extending on . This defines a function from hom to hom given by . Moreover, if hom are distinct, then there is a for which and since is injective, , which implies that the map is injective. Hence, hom hom By a symmetric argument, we have the reverse inequality and so equality holds. 76 Field Theory L' = 1 L E (F) F
Figure 3.2.1 (F) In view of Theorem 3.2.1, we may make the following definition. Definition Let be algebraic and let closed. The cardinality of the set hom over and is denoted by . , where is algebraically is called the separable degree of This new terminology allows us to rephrase the situation for simple extensions. Theorem 3.2.2 (Simple extensions) Let over , with minimal polynomial . Let closed. Then 1) If is separable then and let , where be algebraic is algebraically 2) If is inseparable with radical exponent , then In either case, hom divides . Properties of Separable Degree
Like the ordinary degree, the separable degree is multiplicative. Theorem 3.2.3 If is algebraic then as cardinal numbers. Proof. The set hom of extensions of the inclusion map to an embedding has cardinality . Each extension hom is an embedding of into and can be further extended to an embedding of into . Since the resulting extensions, of which there are , are distinct extensions of to , we have Embeddings and Separability 77 On the other hand, if hom then is the extension of to , hence an element of hom . Since is an extension of to , we see that is obtained by a double extension of and so equality holds in the inequality above. 3.3 Separable Extensions
We have discussed separable elements and separable polynomials. it is now time to discuss separable extensions. Definition An algebraic extension is separable if every element is separable over . Otherwise, it is inseparable. The goal of this chapter is to explore the properties of algebraic extensions with respect to separability. It will be convenient for our present discussion to adopt the following nonstandard (not found in other books) terminology. Definition An algebraic extension is degreewise separable if . An algebraic extension is separably generated if where each is separable over . Simple Extensions
According to Theorem 3.2.2, if only if and , then is separable if and Hence, is separable if and only if is degreewise separable. Moroever, if is degreewise separable, and if , consider the tower The separable degree and the ordinary (vector space) degree are multiplicative and, at least for simple extensions, the separable degree does not exceed the ordinary degree. Hence, implies that the same is true for each step in the tower, and so which shows that is separable over . Thus, extension. Of course, if is separable, then Thus, the following are equivalent: 1) is separable over ; is a separable is separable. 78 Field Theory 2) 3) is degreewise separable; is a separable extension. It is an extremely useful general fact that if then the minimal polynomial of over , that is, min This tells us that if intermediate field . over divides the minimal polynomial of min , then it is also separable over any is separable over In particular, if is separable over field of the form , then it is separable over an intermediate where . But satisfies the polynomial over and so min min divides , which implies that Hence, , or equivalently, . for some , in particular, is separable over . Then Lemma , where and therefore so is For the converse, suppose that 3.1.3 implies that this holds for all is the radical exponent of . But the element . We can now summarize our findings on simple extensions and separability. Theorem 3.3.1 (Simple extensions and separability) Let , with char . The following are equivalent. 1) is separable over . 2) is degreewise separable; that is, be algebraic over 3) is a separable extension. Embeddings and Separability 79 4) There is a for which If in which case for all . is inseparable with radical exponent , then Finite Extensions
Now let us turn to finite extensions . It should come as no surprise that the analogue of Theorem 3.3.1 holds for finite extensions. If is separable, then it is clearly separably generated. If is separably generated by , then , where a finite subset of . Thus, is where is separable over . But is also separable over , since the minimal polynomial of over divides the minimal polynomial of over . Hence, each simple step above is separable and therefore degreewise separable, which implies that is degreewise separable. Finally, if is degreewise separable and , then in the tower the lower step is simple and degreewise separable, hence separable. It follows that is separable over and so is separable. Thus, as in the simple case, separable, separably generated and degreewise separable are equivalent concepts. As to the analogue of part 4) of Theorem 3.3.1, let , where is a finite set. If is separable, then any is separable over and so for any . Thus, , for any . Conversely, if for some , then Lemma 3.1.3 implies that for all . Since is a finite set, we can take to be the maximum of the numbers , where varies over all radical exponents of the elements of , in which case each is separable, and so is separably generated, and therefore separable. 80 Field Theory Theorem 3.3.2 (Finite extensions and separability) Let char be finite. The following are equivalent. 1) is separable. 2) is degreewise separable; that is, . Let 3) is separably generated. 4) If for a finite set which case for all If is not separable, then , then . for some , in for some integer . Algebraic Extensions
For arbitrary algebraic extensions , we have the following. Theorem 3.3.3 (Algebraic extensions and separability) Let char and let be algebraic. 1) is separable if and only if it is separably generated. 2) If is separable and , then for all . Proof. For part 1), if is separable then is separably generated (by itself) over . For the converse, assume that where each is separable over and let . Then for some finite subset . Since is finitely generated and algebraic, it is finite. Thus, Theorem 3.3.2 implies that is separable. Hence is separable over and so is separable. As to part 2), we have for any and which implies that and so . Existence of Primitive Elements
We wish now to describe conditions under which a finite extension is simple. The most famous result along these lines is the theorem of the primitive element, which states that a finite separable extension is simple. We want to state some slightly more general results, and to improve the statements of these results, we need to make some further observations about separable extensions. (These remarks will be repeated and elaborated upon later in the chapter.) Suppose that is a finite extension. Let be the set of all elements of that are separable over . By analogy to algebraic closures, we refer to as the separable closure of in . Note that if , then the extension Embeddings and Separability 81 element of is separably generated and therefore separable. Hence, every is separable over and so is a field. We claim that the extension has no separable elements. For if is separable over , then for the tower we have and so is separable over , which is false. On the other hand, for any , there is a positive integer for which is separable. It follows that and so min divides . Thus, min has only one distinct root. This implies that hom , since any hom must map to itself, for all . Hence, and so We have shown that any finite extension tower can be decomposed into a in which the first step is separable and has the same separable degree as the entire extension. Now we can state our theorem concerning simple extensions. Theorem 3.3.4 1) Any extension of the form where is separable over and is algebraic over is a simple extension. Moreover, if is infinite, this extension has infinitely many primitive elements, of the form where . 2) For any finite extension , there exists a such that If is infinite, there exist infinitely many such elements . 3) (Theorem of the primitive element) If is finite and separable, say where is separable over then is simple. If is infinite, there 82 Field Theory exist infinitely many primitive elements for over of the form where . 4) If has characteristic or if is a finite field then any finite extension of is simple. Proof. If is a finite field, then so is , since is finite. Hence is cyclic and so is simple. Let us now assume that is an infinite field. For part 1), we show that if , with separable over and algebraic over , then , where is algebraic over . The argument can be repeated to obtain a primitive element in the more general case. Let min and min and suppose that the roots of are and the roots of are . Since is separable, the roots of are distinct. However, the roots of need not be distinct. We wish to show that for infinitely many values of , the elements are primitive. To do this, we need only show that , for then . The polynomial has coefficients in and has as a root, and similarily for . Thus, and have the common factor in some extension of . Moreover, since is separable, is a simple root and so no higher power of is a factor of . Therefore, if we can choose so that and have no other common roots in any extension of , it follows that gcd , which must therefore be a polynomial over . In particular, , as desired. The roots of are the values of for which only choose so that none of the roots we need only choose so that and we need satisfy this equation, that is, for and . Part 2) follows from part 1) by considering the separable closure of in . Since is separable, with , we can apply part 1) to the separable extension . Part 3) is a direct consequence of part 1), as is part 4). Example 3.4.1 Consider the extension min . Here we have Embeddings and Separability 83 and min and so and the previous theorem, and . According to is primitive provided that In particular, we can choose any nonzero . Separable Extensions Are Distinguished
We may now establish that the class of separable extensions is distinguished. Theorem 3.3.7 1) The class of separable extensions is distinguished. 2) It is also closed under the taking of arbitrary composites. 3) If is separable, then so is nc , where nc is a normal closure of over . Proof. For the tower property, if the full extension in is separable, then so is . As to , for any , we have min min and so separable over implies separable over . Hence, is separable. Suppose now that and are separable and let . Let be the set of coefficients of min . Then min and so is separable over . It follows that each step in the tower is finite and separable, implying that is separable over . Hence, is separable. For the lifting property, let be separable and let . Since every element of is separable over it is also separable over the larger field . Hence is separably generated and is therefore separable. The fact that separable extensions are closed under the taking of arbitrary composites follows from the finitary property of arbitrary composites. That is, each element of an arbitrary composite involves elements from only a finite number of the fields in the composite and so is an element of a finite composite, which is separable. Finally, a normal closure nc MinPoly and so is generated over is a splitting field in min of the family by the roots of these minimal polynomials, each of 84 Field Theory which is separable over . Hence, nc is therefore separable over . is separably generated over and 3.4 Perfect Fields
Definition A field separable. is perfect if every irreducible polynomial over is It is clear from the definitions that if is perfect then any algebraic extension of is separable. Conversely, suppose that every algebraic extension of is separable. If is irreducible and is a root of in some extension of then is algebraic and so is separable over , that is, is separable. Thus, is perfect. Theorem 3.4.1 A field is separable over . is perfect if and only if every algebraic extension of Theorem 3.4.2 Every field of characteristic and every finite field is perfect. Theorem 3.4.3 Let be a field with char . The following are equivalent. 1) is perfect. 2) for some . 3) The Frobenius map is an automorphism of , for some . If this holds, then 2) and 3) hold for all . Proof. Suppose is perfect. Let and consider the polynomial . If is a root of in a splitting field then and so Now, if is an irreducible factor of separable and so . Thus , that is, Since the reverse inclusion is manifest, we have Lemma 3.1.3. over , then it must be and so . . Then 2) follows from Now assume that 2) holds. Then Lemma 3.1.3 implies that that is irreducible. If is not separable, then . Suppose contradicting the fact that is irreducible. Hence, every irreducible polynomial is separable and so is perfect. Thus, 2) implies 1). Since the Frobenius map is a monomorphism, statement 2), which says that is surjective, is equivalent to statement 3). We can now present an example of a nonperfect field. Embeddings and Separability 85 Example 3.4.2 Let be a prime. Since , the field is perfect. However, if is an independent variable, then the field of all rational functions over is not perfect. We leave proof to the reader. While it is true that any algebraic extension of a perfect field is perfect, not all subfields of a perfect field need be perfect. Theorem 3.4.4 1) If is algebraic and is perfect then is perfect. 2) If is finite and is perfect then is perfect. Proof. Part 1) follows from Theorem 3.4.1 and the fact that every algebraic extension of is an algebraic extension of . For part 2), let char is perfect and and suppose first that is algebraic over . Since . Consider the tower is simple. Thus, is perfect, we have If is the minimal polynomial of over , then and so . It follows that in the tower above, , that is, , whence is perfect. Since is finitely generated by algebraic elements, the result follows by repetition of the previous argument. Note that we cannot drop the finiteness condition in part 2) of the previous theorem since, for example, is algebraic and is perfect even if is not. 3.5 Pure Inseparability
The antithesis of a separable element is a purely inseparable element. Definition An element algebraic over is purely inseparable over if its minimal polynomial min has the form for some . An algebraic extension is purely inseparable if every element of is purely inseparable over . It is clear that for a purely inseparable element , the following are equivalent: (1) is separable, (2) and (3) . In particular, for extensions of fields of characteristic or finite fields, there are no "interesting" purely inseparable elements. 86 Field Theory For , since the coefficient of in min must be a multiple of char , that is, min But min of over . Hence, , where is separable and and we can write min which implies that and . Thus, min where is the radical exponent of over . is , it follows that is the radical exponent , which is separable if and only if Example 3.5.1 Let char . If is transcendental over inseparable over , since its minimal polynomial over . , then is is purely Example 3.5.2 Here we present an example of an element that is neither separable nor purely inseparable over a field . Let char and let be nonzero. Let be transcendental over and let According to Theorem 2.4.5 is algebraic and has degree equal to . Since is a root of the monic polynomial of degree over , this must be the minimal polynomial for over . Since , we deduce that is not separable over . On the other hand, if were purely inseparable over , we would have which would imply that , which is not the case. Hence, separable nor purely inseparable over . Definition Let be finite. Since , we may write is neither where . is the inseparable degree or degree of inseparability of over Embeddings and Separability 87 Note that while the separable degree is defined for infinite extensions, the inseparable degree is defined only for finite extensions. Theorem 3.5.1 Let be a finite extension with char . 1) If then . 2) is separable if and only if . 3) If then , where is the radical exponent of . 4) is purely inseparable if and only if , or equivalently, 5) is a power of . Proof. The first three statements are clear. Part 4) follows from the fact that is purely inseparable if and only if its minimal polynomial has only one distinct root. But this is equivalent to saying that hom has cardinality . Part 5) follows from the fact that is finitely generated and the inseparable degree is multiplicative. We leave the details to the reader. We next characterize purely inseparable elements. Theorem 3.5.2 (Purely inseparable elements) Let char . Let be algebraic over , with radical exponent and let min . The following are equivalent. 1) is purely inseparable over . 2) is a purely inseparable extension 3) for some . Furthermore, is the smallest nonnegative integer for which . Proof. If 1) holds and , then in the tower the inseparable degree of the full extension is equal to the degree, and so the same holds for the lower step. Hence, is purely inseparable over and 2) holds. Clearly, 2) implies 1). If 1) holds, then min holds, then min inseparable. and so , which implies 3). If 3) and, as we have seen, is purely Note that part 3) of the previous theorem, which can be written , is the "antithesis" of the corresponding result for separable. The following result is the analogue of Theorem 3.2.4. Theorem 3.5.3 (Purely inseparable extensions) Let be algebraic. The following are equivalent. 1) is purely inseparably generated; that is, generated by purely inseparable elements. 2) is degreewise purely inseparable, that is, . 88 Field Theory 3) is a purely inseparable extension. Proof. To prove that 1) implies 2), suppose that , where all elements of are purely inseparable over . Any embedding over is uniquely determined by its values on the elements of . But if then is a root of the minimal polynomial min and so . Hence must be the identity and . To show that 2) implies 3), let and suppose that is a root of min in . Then the identity on can be extended to an embedding , for which . Since , we must have and so . Thus, min has only one distinct root in and so is purely inseparable. It is clear that 3) implies 1). Purely Inseparable Extensions Are Distinguished
We can now show that the class of purely inseparable extensions is distinguished. Theorem 3.5.4 The class of purely inseparable extensions is distinguished. It is also closed under the taking of arbitrary composites. Proof. Let . Since pure inseparability is equivalent to degreewise pure inseparability and if and only if and , it is clear that the tower property holds. For lifting, suppose that is purely inseparable and . Since every element of is purely inseparable over , it is also purely inseparable over the larger field . Hence ) is purely inseparably generated and therefore purely inseparable. We leave proof of the last statement to the reader. *3.6 Separable and Purely Inseparable Closures
Let . Recall that the algebraic closure of in is the set of all elements of that are algebraic over . The fact that is a field is a consequence of the fact that an extension that is generated by algebraic elements is algebraic, since if then and so and . We can do exactly the same analysis for separable and purely inseparable elements. To wit, if are separable over , then is separable over . It follows that , and are separable over . Hence, the set of all elements of that are separable over is a subfield of . A similar statement holds for purely inseparable elements. Definition Let
sc . The field separable over for some Embeddings and Separability 89 is called the separable closure of
ic in . The field is purely inseparable over for some is called the purely inseparable closure of we will drop the subscript and write sc and ic in . . When the context is clear, The separable closure allows us to decompose an arbitrary algebraic extension into separable and purely inseparable parts. Theorem 3.6.1 Let be algebraic. sc 1) In the tower the first step is separable and the second step is purely inseparable. 2) Any automorphism of over is uniquely determined by its restriction to sc . Proof. For part 1), if has radical exponent , then has a separable minimal polynomial and is therefore in sc . Thus, Theorem 3.5.2 implies that is purely inseparable over sc . We leave proof of part 2) to the reader. Corollary 3.6.2 Let sc . be finite. Then
sc and Perfect Closures
Let char , where If over and let be an algebraic closure of is perfect. What can we say about ? . Suppose that , then is separable over and therefore cannot be purely inseparable . In other words, the purely inseparable closure ic is contained in . ic On the other hand, we claim that ic is perfect. For if , then . Now, the polynomial has a root in some extension and so ic ic . But then and so . It follows that ic ic ic and so , that is, is perfect. Thus, we have shown that the purely inseparable closure of in is the ic smallest intermediate field that is perfect. This field is also called the perfect closure of in . More on Separable and Inseparable Closures
The remainder of this section is somewhat more technical and may be omitted upon first reading. 90 Field Theory Part 1 of Theorem 3.6.1 shows that any algebraic extension can be decomposed into a separable extension followed by a purely inseparable extension. In ic general, the reverse is not possible. Although is purely inseparable, the ic elements of need not be separable over ic ; they are simply not purely inseparable over . However, it is not hard to see when ic is separable. Theorem 3.6.3 Let be algebraic. Then ic is separable if and only sc ic if . Proof. If ic is separable then so is the lifting sc ic . But since sc sc ic is purely inseparable, so is the lifting sc ic . Thus . sc ic ic sc ic Conversely, if then , being a lifting of a separable sc extension , is also separable. We can do better than the previous theorem when is a normal extension, which includes the case . Let Aut be the set of all automorphisms of over . Since , is also the set of all embeddings of into over . We define the fixed field of in by for all Theorem 3.6.4 Let . Let Aut and let be the fixed field of ic ic in . Then . Furthermore, in the tower , the first step is purely inseparable and the second step is separable. Proof. Let . If is a root of min then there exists an embedding over for which . But and so . ic Hence min has only one root and so . On the other hand, if ic then any must map to itself, since it must map to a root of ic min . Hence . This proves that . and min , . Let where is the set of distinct roots of in . Since any is a permutation of , we deduce that and so the coefficients of lie in . Hence and is separable over . Corollary 3.6.5 If then
ic Now let is separable and sc ic . Let us conclude this section with a characterization of simple algebraic extensions. If is a simple algebraic extension of and if is the radical exponent of , we have seen that is the smallest nonnegative power of such that is separable over , or equivalently, such sc that . It turns out that this property actually characterizes simple algebraic extensions. Before proving this, we give an example where this property fails to hold. Example 3.6.1 Let and Let and be transcendental over with char . It is easily seen that . is purely Embeddings and Separability 91 inseparable with . . However, implies and so Theorem 3.6.6 Let be a finite extension with . Then is simple if and only if is the smallest nonnegative integer for which sc . Proof. We have seen that if is simple then is the smallest such nonnegative integer. For the converse, note first that if is a finite field then so is , implying that is cyclic and so is simple. Let us assume that is sc an infinite field and look at the second step in the tower . This step is purely inseparable. Since sc is finite, we have
sc sc If for some , we have for all , then hypothesis. Hence one of the 's, say , satisfies sc sc sc , contrary to for It follows that
sc sc sc sc sc sc sc sc Since , we have extensions involved are purely inseparable, we get sc Hence, . and since the sc sc . sc sc Our tower now has the form where is purely inseparable sc sc over . In addition, is finite and separable and therefore simple. sc Thus there exists such that sc and the tower takes the form where is separable over and is purely inseparable over . By Theorem 3.3.4, the extension is simple. Note that Theorem 3.6.6 implies that the extension not simple. of Example 3.6.1 is Exercises
1. 2. Find an infinite number of primitive elements for . A biquadratic extension is an extension of degree of the form where and have degree over . Find all the proper intermediate fields of a biquadratic extension. Show that all algebraically closed fields are perfect. If is transcendental over and char , then is not perfect. ic If char and is not perfect, show that . Let be algebraic over , where char and let be the radical exponent of . Show that is separable over if and only if . 3. 4. 5. 6. 92 Field Theory 7. 8. 9. 10. 11. 12. 13. 14. 15. Let and be distinct primes. Then is finite and separable and therefore simple. Describe an infinite class of primitive elements for this extension. Find the minimal polynomial for each primitive element. Let be separable over an infinite field . Prove that there is an infinite number of tuples for which . Show that the class of purely inseparable extensions is closed under the taking of arbitrary composites. sc Prove that for finite, . If is algebraic prove that any automorphism of over is uniquely determined by its restriction to sc . Show that lifting an extension by a purely inseparable extension does not affect the separable degree. That is, show that if is algebraic and is purely inseparable then . Let be finite separable and be finite purely inseparable. Prove that is separable and . In fact, if is a basis for over , prove that it is also a basis for over . Show that if is finite and is finite separable then . Let be a finite extension and let be algebraic over . Let be the set of embeddings of into over . The elements of permute the roots of min . Let be a root of . Show that Hence, the multiset contains copies of each root of . 16. Let be a finite extension that is not separable. Show that for each there exists a subfield of for which is purely inseparable and . 17. Prove that if pcl then the extension pcl is infinite. Chapter 4 Algebraic Independence In this chapter, we discuss the structure of an arbitrary field extension . We will see that for any extension , there exists an intermediate field whose upper step is algebraic and whose lower step is purely transcendental, that is, there is no nontrivial polynomial dependency (over ) among the elements of , and so these elements act as "independent variables" over . Thus, is the field of all rational functions in these variables. 4.1 Dependence Relations
The reader is no doubt familiar with the notion and basic properties of linear independence of vectors, such as the fact that all bases for a vector space have the same cardinality. Independence is a common theme, which applies in the present context as well. However, here we are interested in algebraic independence, rather than linear independence. Briefly, a field element is algebraically independent of a subset if there is no nonconstant polynomial , with coefficients in , for which is a root. Put another way, is algebraically dependent on if is algebraic over . Many of the common properties of linear independence, such as dependence (spanning sets) and bases, have counterparts in the theory of algebraic independence. However, these properties depend only on the most general properties of independence, so it is more "cost effective" to explore these properties in their most general setting, which is the goal of this section. Definition Let be a nonempty set and let relation from to the power set of . We write (read: on ) for , and when for all . dependence relation if it satisfies the following properties, for : be a binary is dependent Then is a all , and 94 Field Theory 1) (reflexivity) 2) (compactness) for some finite subset 3) (transitivity) , 4) (Steinitz exchange axiom) of If we say that is independent of . is dependent if there is an for which is independent. (The empty set is independent.) Definition A subset . Otherwise, The reader should have no trouble supplying a proof for the following lemma. Lemma 4.1.1 1) If then for any superset of . 2) Any superset of a dependent set is dependent. 3) Any subset of an independent set is independent. 4) If is a dependent set, then some finite subset of is dependent. Equivalently, if every finite subset of is independent, then is independent. Theorem 4.1.2 If is independent and then is independent. Proof. Let . If then since , the exchange axiom implies that , a contradiction. Hence . Furthermore, by hypothesis . Thus, is independent. Definition A set is called a base if is independent and . Theorem 4.1.3 Let be a nonempty set with a dependence relation . 1) is a base for if and only if it is a maximal independent set in . 2) is a base for if and only if is minimal with respect to the property that . 3) Let , where is an independent set (possibly empty) and . Then there is a base for such that . Proof. For part 1), assume is a base. Then is independent. If then implies that is not independent, that is, is maximal independent. For the converse, if is a maximal independent set and Algebraic Independence 95 then base. is independent, which is not the case. Hence, and is a For part 2), if is a base, then . Suppose that some proper subset of . If then , contradicting the satisfies independence of . Hence is minimal. Conversely, suppose that is minimal with respect to the condition . If is dependent then for some , a contradiction to the minimality of . Hence is independent and a base for . For part 3), we apply Zorn's lemma. The set of all independent sets in satisfying is nonempty, since . Order by set inclusion. If is a chain in , then the compactness property implies that the union is an independent set, which also lies in . Hence, Zorn's lemma implies the existence of a maximal element , that is, is independent, and is maximal with respect to these two properties. This maximality implies that and so , which implies that is a base. To prove that any two bases for have the same cardinality, we require a lemma, which says that we can remove a particular element from a dependent set and still have a dependent set. Lemma 4.1.4 Let be a finite dependent set and let be an independent subset of . Then there exists for which . Proof. The idea is simply to choose from a maximal independent set containing . In particular, among all subsets of , choose a maximal one for which is independent. Then is a proper (perhaps empty) subset of . If then and so . Theorem 4.1.5 1) If is a finite set for which and if is independent in . 2) Any two bases for a set have the same cardinality. Proof. For part 1), let . Choose . The set satisfies the conditions of the previous lemma (with and so, after renumbering the 's if necessary, we deduce that then ) For any , the set conditions of the lemma (with renumbering, we get satisfies the ) and so, again after possible
2 96 Field Theory Continuing this process, we must exhaust all of the elements of before running out of elements of , for if not, then a proper subset of would have the property that , in contradiction to the independence of . Hence, . Note that this also shows that is finite. If and are bases, we may apply the argument with the roles of and reversed to get . Let us now assume that and are both infinite bases. Thus, . For each , we have and so there is a finite subset such that . This gives a map from to the set of finite subsets of the index set . Moreover, for if then, for any , we have and so , which contradicts the independence of . Hence, Reversing the roles of and shows that . 4.2 Algebraic Dependence
Now that we have the basic theory of dependence, we can return to the subject matter of this book: fields. We recall a definition. Definition Let algebraic over that . . An element is transcendental over , that is, if there is no nonzero polynomial if is not such Recall that if is transcendental over functions in the variable , over the field then . is the field of all rational Definition Let and let . An element is algebraically dependent on over , written , if is algebraic over . If is not algebraically dependent on over , that is, if is transcendental over then is said to be algebraically independent of over and we write . Note that the relation depends on , so we really should write . However, we will not change the base field so there should be no confusion in abbreviating the notation. Algebraic Independence 97 The condition is equivalent to stating that is algebraic. Thus, if and , then if and only if is algebraic for all . But the class of algebraic extensions is closed under arbitrary composites and so this is equivalent to being algebraic. In short, if and only if is algebraic over . Now let us show that algebraic dependence is a dependence relation. Theorem 4.2.1 Algebraic dependence is a dependence relation. Proof. Since any is algebraic over , we have reflexivity: . To show compactness, let and let be the set of coefficients of min . Since each is a rational function over in a finite number of elements of , there is a finite subset of for which . Hence is algebraic over , that is, . For transitivity, suppose that and . Then the tower is algebraic and so is algebraic over , that is, . Finally, we verify the exchange axiom. Suppose that and . Then there is a finite set for which and . Let . Then and . Our goal is to show that , which will follow if we show that . Note that is independent, for if is algebraic over then is algebraic over and so the tower is algebraic, in contradiction to . Thus, according to Lemma 4.1.4, we may remove elements of until the remaining set is independent, and yet is still algebraic over this set. Hence, we may assume that is algebraically independent. Write . If min , then where and are polynomials. Multiplying by the (nonzero) product of the denominators gives 98 Field Theory where and are polynomials over and . Setting gives Now, if the polynomials respect to , then we have and are constant with in contradiction to . Hence, the polynomial is a nonconstant polynomial in over satisfied by , whence as desired. We may now take advantage of the results derived for dependence relations. Definition Let . 1) A subset is algebraically dependent over if there exists that is algebraic over , that is, for which is algebraic. 2) A subset is algebraically independent over if is transcendental over for all . (The empty set is algebraically independent over .) Note that if is algebraic, then certainly is algebraically dependent over , since every is algebraic over , let alone over . The converse, of course, is not true. For example, if is transcendental over then the set is algebraically dependent. In fact, is algebraic over and is algebraic over . However, is far from being algebraic. Algebraic Independence 99 Lemma 4.2.2 1) Any superset of an algebraically dependent set is algebraically dependent. 2) Any subset of an algebraically independent set is algebraically independent. Theorem 4.2.3 If transcendental over is algebraically independent over and then is algebraically independent over is . Algebraic Dependence and Polynomial Relationships
A subset of a vector space is linearly dependent if there is a nontrivial linear relationship among the vectors of . A similar statement holds in the present context. Definition Let . A subset has a nontrivial polynomial relationship over if there is a nonzero polynomial over for which , for distinct . This is equivalent to saying that some is algebraic over the ring of polynomials in . To see that the two statements in the definition are equivalent, suppose that for distinct , where is a nonzero polynomial over . If , then this simply says that if and only if is algebraic over . For , we may assume that do not enjoy a similar polynomial dependency and hence that where and . Then the nonzero polynomial satisfies , showing that is algebraic over . Now, to say that is algebraically dependent is to say that is algebraic over for some . This is to say that is algebraic over the field of rational functions in . But this is equivalent to saying that is algebraic over the ring of polynomials in . One direction is clear, since a polynomial is a rational function. On the other hand, if satisfies a polynomial of degree over then has the form where . Multiplying by the product of the denominators gives a polynomial satisfied by and whose leading coefficient is not zero. 100 Field Theory We have proved the following. Theorem 4.2.4 Let . A subset of is algebraically dependent over if and only if there is a nontrivial polynomial relationship in . 4.3 Transcendence Bases
We can now define an analogue of a (linear) basis for a vector space. Definition Let . A transcendence basis for over is a subset that is algebraically independent over and for which algebraic. is Since algebraic dependence is a dependence relation, we immediately get the following two results. Theorem 4.3.1 Let . A subset is a transcendence basis for over if and only if it satisfies either one of the following. 1) is a maximal algebraically independent subset of over . 2) is a minimal set satisfying , that is, is minimal for the property that is algebraic. Theorem 4.3.2 Let . 1) Any two transcendence bases for over have the same cardinality, called the transcendence degree of over and denoted by . 2) Suppose where is algebraically independent over and is algebraic. Then there exists a transcendence basis for over satisfying . In particular, . While the vector space dimension is multiplicative over a tower of fields, the transcendence degree is additive, as we see in the next theorem. Theorem 4.3.3 Let . 1) If is algebraically independent over and is algebraically independent over then is algebraically independent over . 2) If is a transcendence basis for over and is a transcendence basis for over then is a transcendence basis for over . 3) Transcendence degree is additive, that is, Proof. For part 1), consider a polynomial dependence of a polynomial over , that is, for which , where are distinct and are Algebraic Independence 101 distinct. Write where and where the monomials such monomial, the monomials polynomial are distinct and, for each are distinct. Consider the over . Since is algebraically independent over , it follows that However, is algebraically independent over and so the zero polynomial and is algebraically independent over . Thus, . is For part 2), we know by part 1) that is algebraically independent over . Also, since and are algebraic, each step in the tower is algebraic and so is algebraic. Hence, is a transcendence basis for over . Part 3) follows directly from part 2). Purely Transcendental Extensions
When one speaks of the field of rational functions in the "independent" variables , one is really saying that the set is algebraically independent over and that . We have a name for such an extension. Definition An extension is said to be purely transcendental if for some transcendence basis for over . We remark that if is purely transcendental over then for some transcendence basis , but not all transcendence bases for over need satisfy . The reader is asked to supply an example in the exercises. The following is an example of an extension that is neither algebraic nor purely transcendental. Example 4.3.1 Let transcendental over and let be a field with char . Let be be a root of in some splitting , let 102 Field Theory field and let . Clearly, is not algebraic over . We contend that is also not purely transcendental over . Since is algebraically independent and is algebraic, the set is a transcendence basis for over and so . If were purely transcendental over there would exist a transcendental element over for which . Let us show that this is not possible. If then and where and are polynomials over . Hence or This can be written for nonconstant polynomials and , which we may assume to be pairwise relatively prime. Let us assume that deg deg , in which case deg deg . We now divide by and take the derivative with respect to to get (after some simplification) Since and implies are relatively prime, we deduce that deg deg . Hence, deg deg . But this deg which is not possible for transcendental. is not purely Purely transcendental extensions are 100% transcendental, that is, every element of is transcendental over . Theorem 4.3.4 A purely transcendental extension is 100% transcendental, that is, any is transcendental over . Proof. Let be a transcendence basis for over . Since , it follows that for some finite set , and we can assume that . Letting , we have where is a simple transcendental extension of . Algebraic Independence 103 Hence, Theorem 2.4.5 implies that also over . is transcendental over , and therefore The following result will prepare the way to finishing the proof (promised in Chapter 2) that the class of finitely generated extensions is distinguished. Theorem 4.3.5 Let and suppose that is algebraic. If is algebraically independent over , then is also algebraically independent over . In other words, remains algebraically independent over any algebraic extension of the base field. Proof. We have the picture shown in Figure 4.3.1 K(T)
algebraic K
algebraic F(T) F
Figure 4.3.1 Since is algebraic , so is the lifting . Now, and so which shows that must be a transcendence basis for over . For an alternative proof, if is not algebraically independent over , there exists that is algebraic over . Since is algebraic, the lifting is also algebraic, and so the tower is algebraic, whence is algebraic over algebraic independence of over . , in contradiction to the Finitely Generated Extensions Are Distinguished
We are now in a position to finish the proof that the class of finitely generated extensions is distinguished. Note how much more involved this task is than showing that finite or algebraic extensions are distinguished. 104 Field Theory Theorem 4.3.6 Let . If is finitely generated over then is also finitely generated over . Thus, the set of finitely generated extensions is distinguished. Proof. Let be a transcendence basis for over . Then the second step in the tower is algebraic and is finitely generated over . Hence, if we can prove the theorem for algebraic intermediate fields, we will know that is finitely generated over and therefore also over , since is a finite set. Thus, we may assume that is finite. Let Our plan is to show that with algebraic and show that be a transcendence basis for over . (see Figure 4.3.2) by showing that any finite subset of that is linearly independent over is also linearly independent over , as a subset of . Since is finitely generated and algebraic, is finite and the proof will be complete. E
finite K
algebraic F(T) F
Figure 4.3.2 Let be linearly independent over and suppose that where . We wish to show that . By clearing denominators if necessary, we may assume that each is a polynomial over . Collecting terms involving like powers of the 's gives
, where is the coefficient of in . Since , Theorem 4.3.5 implies that is algebraically independent over , it follows that does not satisfy any polynomial relationships over and so Algebraic Independence 105 , Then the linear independence of over
, gives and so over for all . This shows that , as desired. is linearly independent *4.4 Simple Transcendental Extensions
The class of purely transcendental extensions is much less well behaved than the class of algebraic extensions. For example, let be transcendental over . Then 2 in the tower , the extension is purely 2 transcendental (and simple) but the second step is not transcendental at all. In addition, if is purely transcendental and , it does not necessarily follow that the first step is purely transcendental. However, this is true for simple transcendental extensions. The proof of this simple statement illustrates some of the apparent complexities in dealing with transcendental extensions. Theorem 4.4.1 (Luroth's Theorem) Let be transcendental over . If and then for some . Proof. Let us recall a few facts from Theorem 2.4.5. Since , Theorem 2.4.5 implies that for any , the tower is algebraic. Theorem 2.4.5 also implies that if where and are relatively prime polynomials over , then max deg Now, we want to find an showing that . Of course, Let min where are relatively prime. Since is not algebraic over , not all of the coefficients of can lie in . We will show that for any coefficient , we may take deg , . for which To this end, consider the polynomial 106 Field Theory Since , we have other words, there exists . But such that and so over . In or Multiplying both sides of this by gives (4.4.1) where Now let be the greatest common divisor of the coefficients on the righthand side of this equation. Since divides the first coefficient , it must be relatively prime to each and so for all . Factoring out gives where , , is primitive in , that is, nonconstant polynomial in . Note, however, that for each , the polynomial . Also, for each , we have is not divisible by any appears in the coefficent of It follows that the degree of deg , , with respect to satisfies deg (4.4.2) max deg Thus, (4.4.1) can be written Algebraic Independence 107 (4.4.3) Next, we multiply both sides of (4.4.3) by a polynomial the denominators of , giving that will clear all of where divide . Since is primitive, we must have and so the other factor must , that is, there exists a polynomial for which (4.4.4) Now, the degree of the lefthand side of this equation is at most max deg deg and by (4.4.2), the degree of the righthand side is at least . Hence, the degree of either side of (4.4.4) is and (4.4.2) implies that deg , that is, (4.4.5) where . Since the right side of (4.4.5) is not divisible by any nonconstant polynomial in , neither is the left side. But the left side is symmetric in and , so it cannot be divisible by any nonconstant polynomial in either. Hence, is not divisible by any nonconstant polynomial in , implying that , that is, (4.4.6) where . Finally, since the degree and degree of the left side of (4.4.6) agree, this is also true of the right side. Hence by (4.4.2), deg Thus, deg , and the proof is complete. It can be shown that Luroth's theorem does not extend beyond simple transcendental extensions, but a further discussion of this topic would go beyond the intended scope of this book. The Automorphims of a Simple Transcendental Extension
We conclude with a description of all automorphisms of a simple transcendental extension . Let GL denote the general linear group, that is, the group of all nonsingular matrices over . The proof, which is left as an exercise, provides a nice application of Theorem 2.4.5. 108 Field Theory Theorem 4.4.2 Let be a simple transcendental extension and let Aut denote the group of all automorphisms of over . 1) For each which GL there is a unique Aut for Moreover, all automorphisms of GL . 2) If GL , then over have the form for some and Also, if and only if is a nonzero scalar matrix. In other words, the map GL Aut defined by is an epimorphism with kernel equal to the group of all nonzero scalar matrices in GL . Exercises
1. Find an example of a purely transcendental extension with two transcendence bases and such that but is a proper subfield of . 2. Let and . Show that . 3. Let and let . Show that with equality if is algebraically independent over or algebraic over . 4. Use the results of the previous exercise to show that if and then . 5. Let be a field of characteristic and let be transcendental over . Suppose that . Show that is a purely transcendental extension by showing that where . 6. Show that the extension , where is transcendental over , is not purely transcendental. 7. Let and suppose that is algebraically independent over . Prove that is algebraic if and only if is algebraic. 8. Prove that the transcendence degree of over is uncountable. 9. a) Show that the only automorphism of is the identity. b) Show that the only automorphisms of over are the identity and complex conjugation. c) Show that there are infinitely many automorphisms of over . 10. (An extension of Luroth's theorem) Suppose that is purely transcendental. Show that any simple extension of contained in (but not equal to ) is transcendental over . Algebraic Independence 109 11. Prove part 1) of Theorem 4.3.5 by contradiction as follows. Suppose that is algebraically dependent over . Then there exists an that is algebraic over for some finite sets and not containing , and we may assume that no proper subset of has the property that is algebraic over . Prove that . Prove that is not empty. If , prove that is algebraic over . Complete the proof from here. 12. Prove Theorem 4.4.2. Part IIGalois Theory Chapter 5 Galois Theory I: An Historical Perspective Galois theory sits atop a structure of work began about 4000 years ago on the question of how to solve polynomial equations algebraically by radicals, that is, how to solve equations of the form by applying the four basic arithmetical operations (addition, subtraction, multiplication and division), and the taking of roots, to the coefficients of the equation and to other "known" quantities (such as elements of the base field). More specifically, a polynomial equation there is a tower of fields is solvable by radicals if where contains a splitting field for (and hence a full set of roots of ) and where each field in the tower is obtained by adjoining some root of an element of the previous field, that is, where . In this chapter, we will review this structure of work from its beginnings in Babylonia through the work of Galois. In subsequent chapters, we will set down the modern version of the theory that has become known as Galois theory. 5.1 The Quadratic Equation
Archeological findings indicate that as early as about 2000 B.C., the Babylonians (Mesopotamians) had an algorithm for finding two numbers and whose sum and product were known. The algorithm is 1) Take half of . 2) Square the result. 114 Field Theory 3) Subtract . 4) Take the square root of this result. 5) Add half of . This results in one of the values and : the other is easily obtained. This algorithm can be expressed in modern notation by the formula for solving the system of equations The solutions to this system are solutions to the quadratic equation Thus, except for one issue, it can be said that the Babylonians knew the quadratic formula, but in algorithmic form. The one issue is that the Babylonians had no notion of negative numbers! Indeed, they developed a separate algorithm to compute the numbers and whose difference and product were known. This is the solution to the system whose solutions satisfy the quadratic equation Unfortunately, the origin of the Babylonian algorithms appears lost to antiquity. No texts uncovered from that period indicate who or how the algorithm was developed. 5.2 The Cubic and Quartic Equations
In the 3500 years or so between the apparent achievement of the Babylonians and the midRenaissance period of the 1500's, not much happened in Europe of a mathematical nature. However, during the Middle Ages (that is, prior to the Renaissance, which began in the late thirteenth century), the Europeans did learn about algebra from the Arabs and began to devise a new mathematical symbolism, which opened the way for the dramatic advancements of the midRenaissance period. In particular, solutions to the general cubic and quartic equations were discovered. As to the cubic, we have the following excerpt from Girolamo Cardano's Ars Magna (1545). (Cardano was a highly educated and skilled physician, natural philosopher, mathematician and astrologer.) Galois Theory I: An Historical Perspective 115 In our own days Scipione del Ferro of Bologna has solved the case of the cube and first power equal to a constant, a very elegant and admirable accomplishment. Since this art surpasses all human subtlety and the perspicuity of mortal talent, and is a truly celestial gift and a very clear test of the capacity of men's minds, whoever applies himself to it will believe that there is nothing that he cannot understand. In emulation of him, my friend Niccol Tartaglia of Brescia, wanting not to be outdone, solved the same case when he got into a contest with his [Scipione's] pupil, Antonio Maria Fior, and, moved by my many entreaties, gave it to me. The solution of the quartic equation was discovered by one of Cardano's students, Ludivico Ferrari, and published by Cardano. Let us briefly review these solutions in modern notation. Solving the Cubic
1) An arbitrary monic cubic polynomial form
3 3 can be put in the by replacing by 2) Introduce the variables . and and set . Then has the form or, equivalently, 3) If , then we get Thus, a solution to the pair of equations provides a solution second equation by to the original cubic equation. Multiplying the and using the fact that gives which is a quadratic equation in . If quadratic, then , so that is a cube root of a solution to this is a root of the original cubic. 116 Field Theory Solving the Quartic
1) An arbitrary monic quartic equation can be put in the form 2) Introducing a variable , we have Using the quartic equation from 1) to replace on the right, we have 3) If the right side of this equation can be put in the form , then we can take square roots. This happens if the quadratic on the right has a single root, which happens if its discriminant is , that is, if which is a cubic in , and can therefore be solved, as described earlier. 4) Once is found, we have and , and so our quartic is Hence, which can be solved for a solution of the original quartic. 5.3 HigherDegree Equations
Naturally, solutions to the arbitrary cubic and quartic equations led to a search for methods of solution to higherdegree equations, but in vain. It was not until the 1820s, some 300 years later, in the work of Ruffini, Abel and then Galois, that it was shown that no solution similar to those of the cubic and quartic equations could be found, since none exists. Specifically, for any , there is no algebraic formula, involving only the four basic arithmetic operations and the taking of roots, that gives the solutions to any polynomial equation of degree . In fact, there are individual quintic (and higherdegree) equations whose solutions are not obtainable by these means. Thus, not only is there no general formula, but there are cases in which there is no specific formula. Galois Theory I: An Historical Perspective 117 5.4 Newton's Contribution: Symmetric Polynomials
It was not until the accomplishments of Vandermonde and, to a larger extent, Lagrange, in the period around 1770, that a deeper understanding of the work that led to the solutions of the cubic and quartic equations was revealed. However, even these fine mathematicians were unable to take the leap made by Abel and Galois a few decades later. The cornerstone of the work of Vandermonde and Lagrange is the work of Isaac Newton on symmetric polynomials. We will go into precise detail at the appropriate time in a subsequent chapter, but here is an overview of Newton's contribution in this area. The Generic Polynomial
If are independent variables, the polynomial is referred to as a generic polynomial of degree . (Galois would have referred to this as a polynomial with "literal" coefficients.) Since the roots of the generic polynomial are independent, this polynomial is, in some sense, the most "general" polynomial of degree and facts we learn about often apply to all polynomials. It can be shown by induction that the generic polynomial can be written in the form where the coefficients are given by These polynomials are called the elementary symmetric polynomials in the variables . Thus, except for sign, the coefficients of are the elementary symmetric polynomials of the roots of . Moreover, since this holds for the generic polynomial, it is clear that it holds for all polynomials. Symmetric Polynomials
Intuitively, a polynomial in the variables is symmetric if it remains unchanged when we permute the variables. More carefully, is symmetric if 118 Field Theory for any permutation of . Of course, each elementary symmetric polynomial , that is, each coefficient of , is a symmetric polynomial of the roots , in this sense. It follows that any polynomial (symmetric or otherwise) in the coefficients of is a symmetric polynomial of the roots . For instance, is unchanged by a permutation of the 's. Isaac Newton realized, sometime in the late 1600s, that a kind of converse to this holds: Any symmetric polynomial in the roots of is a polynomial in the coefficients of . Let us state this theorem, known as Newton's theorem, first without reference to roots. Newton's Theorem 1) A polynomial is symmetric if and only if it is a polynomial in the elementary symmetric functions , that is, Moreover, if has integer coefficients, then so does . 2) Let be a polynomial. Then the set of symmetric polynomials in the roots of is the same as the set of polynomials in the coefficients of . In particular, any symmetric polynomial in the roots of belongs to the same field as the coefficients, so if is a polynomial over , then any symmetric polynomial in the roots of belongs to . Also, if has integer coefficients, then any symmetric polynomial in the roots of is an integer. The proof of Newton's theorem will be given in a later chapter. However, it should be noted that the proof is in the form of an algorithm (however impractical) for finding the polynomial . How can this be used to advantage in the present context? The answer is both simple and profound: When trying to find the roots of a polynomial , we can assume not only that the coefficients of are known (obviously), but also that any symmetric polynomial in the roots of is known! The reason is that an algorithm is known for computing this symmetric polynomial of the roots that requires knowledge of the coefficients of the polynomial only (and of other known quantities, such as rational numbers). Galois Theory I: An Historical Perspective 119 For instance, if has roots and , then not only are and known, being the coefficients of , up to sign, but we may also assume that expressions such as are known. More to the point, we cannot assume that is known, but we can assume that is known and so we may write where gives and are known. Hence, , or . Adding this to the first equation Of course, and becomes the wellknown quadratic formula and so this (Note that there is another solution to , which gives the other root.) We are very close here to the work of Vandermonde and Lagrange. 5.5 Vandermonde
How can we apply the previous analysis to the cubic equation? The previous solution to the quadratic can be expressed as where the solutions are and . Now let and be solutions to a cubic equation. Again, the sum is known, being symmetric in the roots. As to the analogue of the difference, note that the coefficients and of are the two roots of the equation , that is, they are the square roots of unity. In general, the complex of the equation th roots of unity are the roots (in the complex field) As we will see in a later chapter, this equation has distinct complex roots, which we denote by . The set is a cyclic group under multiplication. Any generator of is called a primitive th root of unity. The set of primitive th roots of unity is denoted by . Note that if , then 120 Field Theory This fact is used many times. Now, for the analogue of the difference , we require the two expressions where is a primitive cube root of unity. Then Now, the expressions and are not symmetric in the roots, so we cannot conclude directly from Newton's theorem that they are known. However, the previous expression can be written in the form and while the expressions are also not symmetric in the roots, the expressions and are symmetric. To see this, first note that interchanging and has the effect of interchanging and , thus preserving both and . Also, the cyclic permutation , which sends to , to and to , actually fixes both and . For example, Thus, both and are known quantities, from which we can compute and using the quadratic formula. It follows that the root is known. Note that there are three possible values for each cube root in this expression, leading to nine possible value of , of which exactly are roots of the cubic. Of course, it is a simple matter (in theory) to determine which of the Galois Theory I: An Historical Perspective 121 nine candidates are roots. Thus, the solution to the general cubic equation is reduced to solving a quadratic equation and to the taking of cube roots. This analysis of the cubic equation is the work of Vandermonde, who presented it to the Paris Academy in 1770, along with a similar analysis of the quartic and some additional work on higherdegree polynomial equations. However, Vandermonde appears not to have pursued this work beyond this point. Perhaps we can find one reason in the fact that Lagrange's major (over 200 page) treatise Rflexions sur la Rsolution Algbrique des Equations, which included similar but independent work in more depth on this subject, was published a few months after Vandermonde's presentation, while Vandermonde had to wait until 1774 to see his work published by the Paris Academy! 5.6 Lagrange
In his Rflexions, Lagrange gives a thorough treatment of the quadratic, cubic and quartic equations. His approach is essentially the same as Vandermonde's, but with a somewhat different perspective. He also addresses some issues that Vandermonde did not. The Cubic Equation
Lagrange also considers the expression but looks directly at all six quantities obtained from this expression by substituting the roots and : The roots of are given in terms of the 's and other known quantities by Note that, in the notation of the previous section, and . 122 Field Theory Now, permuting the roots in any of the coefficients of the th degree polynomial 's results in another and so the are symmetric in the 's and therefore also symmetric in the roots and are therefore known quantities. Lagrange called the equation resolvent equation and the solutions to this equation resolvents. and , the Lagrange observed that although the resolvent equation is of degree , it is also a quadratic equation in , due to the relationships among the 's. In particular, can be expressed in terms of and only: Thus, the resolvent equation is easily solved for the six resolvents , using the quadratic formula, followed by the taking of cube rootsthe same operations required by Vandermonde's approach. It is then a matter of determining which roots correspond to and . Lagrange addresses (or avoids) the latter issue by observing that if resolvent, we can assume, by renaming the roots and , that since it is easily checked that the product is symmetric in and therefore known, the three roots of are given by is any . Then and Thus, the solutions to the cubic are expressed in terms of any resolvent. The important points to note here are that 1) Each resolvent is an expression (polynomial) in the roots of and other known quantities. 2) Conversely, the roots of can be expressed in terms of a single resolvent and other known quantities. 3) Each resolvent can be determined in a tractable way, in this case by solving a quadratic equation and taking cube roots. Galois Theory I: An Historical Perspective 123 The Quartic Equation
Lagrange and Vandermonde each employed their similar lines of analysis with success for quartic equations. For a quartic , the resolvent expression is where the 's represent the roots of and where is a primitive th root of unity. It follows that there are distinct resolvents, satisfying a resolvent equation of degree . By analogy with the cubic case, one root of the quartic is given by since and each appear in all three of the last positions in and so have coefficient . It is possible to proceed in a manner analogous to the cubic case, but Lagrange and Vandermonde both observed that a simplification is possible for the quartic. In particular, unlike the case of the cubic (and the quintic), where the degrees are prime, in the case of a quartic, there is a nonprimitive th root of unity other than , namely, . The resolvent expression with respect to , has only distinct resolvents, which have the form Moreover, the roots of are given by and . Since the resolvent polynomial in this case is the resolvent equation, whose coefficients are known, can be solved by solving 124 Field Theory a known cubic equation . This gives solutions and , leaving only an ambiguity of sign in determining the resolvents and . Lagrange addressed the issue of how to choose the correct sign, but Vandermonde simply left the issue to one of trial and error. The Quintic Equation
The case of the th degree equation stymied both mathematicians, and for good reason. The Lagrange resolvent equation has degree and is a th degree equation in . It seems that both mathematicians doubted that their lines of analysis would continue to be fruitful. The somewhat ad hoc trick used for the quartic will not work for the quintic, and it is clear that the Lagrange Vandermonde resolvent approach is simply running out of steam. This is essentially where Lagrange (and Vandermonde) left the situation in his Rflexions. 5.7 Gauss
We need to say a word about roots of unity with respect to solvability by radicals. It is an obvious fact that since we allow the taking of roots in constructing a tower that shows that is solvable by radicals, then every equation of the form is solvable by radicals, that is, the th roots of unity are obtainable by takingwellroots. This is not a very useful statement. Note, however, that if polynomial is an th root of unity, then is a root of the which has degree . It would be much more interesting (and useful) to know that could be obtained by adjoining roots whose degree is at most , that is, various th roots, where . This was Gauss's contribution, published in 1801 in his Disquisitiones Arithmeticae, when he was only 24 years old. We should mention that while Gauss is considered by many to be perhaps the greatest mathematician of all time, in this particular case, the ideas that Gauss used appear not to have originated with him. Moreover, Gauss seems to leave a gap in his proof, so one could argue that this was not really a completely Gaussian affair. Let us briefly outline Gauss's approach, which uses Lagrange resolvents. First, it is not hard to show that if , where and are relatively prime, then every primitive th root of unity is the product of a primitive th root of unity and a primitive th root of unity. In symbols, Galois Theory I: An Historical Perspective 125 Moreover, since we have (proof postponed until a later chapter), where , and so we need to prove the result only for th roots of unity, where prime. A primitive th root of unity is a solution to the polynomial equation is a whose solutions are These are all primitive th roots of unity, since is prime. Note that the exponents of constitute the cyclic group of nonzero elements of the field . Any generator of this group is called a primitive root modulo . For any such , we have (5.7.1) Now, since the equation has degree , a Lagrange resolvent for this equation requires a primitive st root of unity , and the resolvent expression is where, as usual, a resolvent is obtained by substituting the roots of 's. for the The key idea (which may have been in part Vandermonde's) is to choose a resolvent in a specific way. In particular, the roots are chosen in the order given by a primitive root modulo , as shown in (5.7.1). Hence, the resolvent is Note that for any , Accordingly, if we take the sum the coefficient of will be , for all . Also, the coefficient of is 126 Field Theory and so , that is, Thus, if it can be shown that the expressions under the radical signs are known, then will be known, at least up to determining the correct st roots. This is where the order of the roots in the resolvent is important. (Actually, the issue of which roots to take can be mitigated considerably, but we will not go into the details here.) The "hard part" is thus to show that the expressions are known. Since we can assume that is known (being a smaller primitive root of unity), it suffices to show that does not depend on . This is done using a result whose origin is somewhat obscure. Gauss apparently used the result without proof at one point and then later gave an incomplete proof. In any case, it is not entirely clear whether Gauss possessed a complete proof of this result, which can be stated as follows. Theorem 5.7.1 Let be a primitive th root of unity and let st root of unity. Then the powers be a primitive are linearly independent over . Proof. We need the following additional facts about roots of unity, whose proofs will be given in a later chapter. 1) If is a primitive th root of unity, then , where is the Euler phi function, that is, is the number of positive integers less than and relatively prime to . 2) If and are relatively prime, then . 3) If is a prime, then . Consider the tower The lower step has degree , has degree and the upper step, being a lifting of . Consider also the tower The lower step has degree has degree , then and if the upper step Galois Theory I: An Historical Perspective 127 Hence, , which implies that and , that is, Hence, the set is a basis for over . Now let us look at how this result can be used to show that does not depend on modulo ), we have . If we replace by (recall that is a primitive root It follows that In other words, is invariant under the replacement . (which are Now, is a polynomial in and . Collecting powers of linearly independent by Theorem 5.7.1) gives Then the invariance under implies that Equating coefficients of the linearly independent powers of gives and so the polynomial expressions , for , are equal. Hence, 128 Field Theory which is independent of , as desired. Thus, we have shown that a primitive th root of unity can be expressed in terms of a primitive st root of unity , using only root of degree at most . An induction completes the proof that any th root of unity can be expressed by taking roots of degree at most . As a very simple illustration, let us compute a primitive cube root of unity . We begin with a primitive square root of unity and form the expressions Then since , we have Thus, and we need only choose the correct combination of signs. 5.8 Back to Lagrange
As we have remarked, Lagrange's (and Vandermonde's) resolvent has three properties: 1) Each resolvent is a polynomial in the roots of and other known quantities, including perhaps the th roots of unity. 2) Conversely, the roots of can be expressed in terms of a single resolvent and other known quantities. 3) Each resolvent can be determined in a tractable way. Lagrange doubted that it would be possible to find a resolvent that could be determined in a tractable way for the quintic, let alone for higherdegree polynomials. On the other hand, he did spend considerable effort considering "resolvents" that satisfy only 1) and 2). In fact, the following theorem of Lagrange, and its corollary, is a cornerstone of Galois theory. The version we present here appears in Edwards, and is from Lagrange's Rflexions, Article 104. Theorem 5.8.1 If of and are any two functions [polynomials] in the roots and if these functions Galois Theory I: An Historical Perspective 129 are such that every permutation of the roots which changes also changes , one can, generally speaking, express rationally in terms of and so that when one knows a value of one will also know immediately the corresponding value of ; we say generally speaking because if the known value of is a double or triple or higher root of the equation for then the corresponding value of will depend on an equation of degree or or higher with coefficients that are rational in and . If we think of as a known polynomial of the roots, then this theorem states that under the conditions of the theorem, the value of , which could simply be a root of , is expressible as a known function of . Lagrange's theorem has the following corollary (in slightly more modern notation). Corollary 5.8.2 Suppose that exists a polynomial has distinct roots, say, with the property that the . If there values are distinct, that is, if is changed by every permutation of the roots, then any polynomial in the roots, including the roots themselves, is a known rational expression in . We will be able to rephrase this in more modern terms in a later chapter. For the curious, it is as follows: If is separable over , with splitting field and Galois group and if has the property that for all , then and so taking fixed fields gives , that is, every polynomial in the roots of is a polynomial in . A polynomial as described in the previous corollary is a "resolvent" in the sense that it satisfies the first two conditions of a Lagrange resolvent: is a known function of the (unknown) roots and the roots are a known function of . Any with these properties is called a Galois resolvent, because Galois was the first to recognize that such a resolvent always exists (provided that has no multiple roots). He was also the first to realize the importance of such resolvents. We can describe Galois resolvents in more be a splitting field for is the field of "known" quantities. Then if , that is, if and only if modern terms as follows. Let over . We may assume that is a Galois resolvent if and only is a primitive element of . Now we see that the existence of Galois resolvents follows from the Theorem of the Primitive Element. Assuming that has no multiple rootsan assumption that Galois also madethe fact that is finite and separable implies that it is simple. 130 Field Theory 5.9 Galois
It is not hard to place the work of Evariste Galois in time, since he was born in 1811 and died only 21 years later, of a gunshot wound, in 1832. However, it is much harder to describe the importance of his work, which sparked the foundations of modern algebra. (Of course, Cauchy, Cayley, Lagrange, Vandermonde, Newton, Gauss and others had a hand in the foundations of algebra as well.) Galois realized that while a (Galois) resolvent might not be able to provide the actual values of the roots of a polynomial, it does lead the way to a beautiful theory, now called Galois theory that, among other things, shows that there are no Lagrange resolvents for polynomials of degree or greater. In his 1831 Memoir on the Conditions for Solvability of Equations by Radicals, Galois states a result akin to the corollary of Lagrange given above, without mention of either Lagrange or his theorem (although he had read Lagrange as a student). Moreover, Galois' proof is, to say the least, sketchy. In fact, when Poisson read Galois' memoir, as submitted for publication to the Paris Academy of Sciences, Poisson remarked "We have made every effort to understand Mr. Galois' proof. His arguments are not clear enough, nor developed enough, for us to be able to judge their correctness ." Galois' paper was rejected for publication. In his memoir of 1831, Galois proved the following result (Proposition VIII): "For an equation of prime degree, which has no commensurable divisors, to be solvable by radicals, it is necessary and sufficient that all roots be rational functions of any two of them." In more modern language, this theorem says that if is irreducible and separable of prime degree , then the equation is solvable by radicals if and only if is a splitting field for , for any two roots and of . Since, for example, any quintic polynomial with exactly two nonreal roots fails to meet this condition, it cannot be solvable by radicals. This theorem is covered in detail in the chapter on solvable extensions. Galois and Groups
Galois' great achievement was not the actual result that polynomial equations of degree and higher have no general algebraic solution. Indeed, even the formulas for cubic and quartic equations are not of much practical use. Galois' great achievement lies in the path he took to prove this result, in particular, his Galois Theory I: An Historical Perspective 131 discovery and application of the notion of a "Galoisstyle" group, described below. While on the subject of groups, it cannot be said that Galois discovered in its entirety the modern notion of a group. As we will see, Galois dealt only with sets of permutations and stated only that these sets must be closed under composition (although not in these words). The other properties of the definition of a modern group: associativity, identity and inverses, were not mentioned explicitly by Galois. (Perhaps he thought them too obvious for explicit mention.) When Galois' work was finally published in 1846, the theory of finite permutation groups had already been formalized by Cauchy, who likewise required only closure under product, but who clearly recognized the importance of the other axioms by introducing notations for the identity and for inverses. Cayley (1854) was the first to consider the possibility of more abstract groups, and the need to axiomatize associativity. He also axiomatized the identity property, but still assumed that each group was a finite set, and so had no need to axiomatize inverses (only the validity of cancellation). It was not until 1883 that Dyck, in studying the relationship between groups and geometry, made explicit mention of inverses. It is also interesting to note that Cayley's famous theorem of group theory, to the effect that every group is isomorphic to a permutation group, completes a full circle back to Galois (at least for finite groups)! GaloisStyle Groups
Galois' version of a group is as follows (although the terminology is not necessarily that of Galois). Consider a table in which each row contains an ordered arrangement of a set of distinct symbols (such as the roots of a polynomial), for example Then each pair of rows defines a permutation of , that is, a bijective function on . Galois considered tables of ordered arrangements with the property that the set of permutations that transform any given row into the other rows (or into itself) is the same for all rows , that is, for all . Let us refer to this type of table, or list of ordered arrangements, as a Galoisstyle group. 132 Field Theory It is not hard to show that a list of arrangements is a Galoisstyle group if and only if the corresponding set ( ) of permutations is a subgroup of the group of all permutations of the set , that is, if and only if is a permutation group, in the modern sense. To see this, let the permutation that transforms row to row be . Then Galois' assumption is that the sets are the same for all . This implies that for each and , there is a for which . Hence, and so is closed under composition. It is also closed under inverses, since for any , it is true that . Finally, the identity is in , since it is the substitution associated to the pair of rows . Conversely, if is a permutation group, then since it follows that for all . Galois appears not to be entirely clear about a precise meaning of the term group, but for the most part, he uses the term for what we are calling a Galois style group. Galois also worked with subgroups and recognized the importance of what we now call normal subgroups, although his "definition" is quite different from what we would see today. The Galois Group
For a modern mathematician, the Galois group of a polynomial over a field is defined in terms of a splitting field. Galois and his predecessors talked about the "roots" of a polynomial without regard to considerations of their existence (much as our students do today) and it was not until Kronecker came upon the scene, several decades later, that the issue of existence was explicitly addressed. In any case, the modern definition of the Galois group of a polynomial over is the group of all automorphisms of a splitting field of over that fix pointwise, in symbols Aut Galois would have defined the Galoisstyle group of a polynomial distinct roots, essentially as follows (but in different terms). Let splitting field for . Let be the minimal polynomial of over , with be a and let be the conjugates of , that is, the roots of . Note that since is assumed Galois Theory I: An Historical Perspective 133 to have only simple roots, the extension is separable and so separable, that is, has distinct roots. Also, since is normal, splits over and so . is Let be the roots of . Each root is a polynomial in the primitive element (that is, the Galois resolvent). Consider the list of arrangements (5.9.1) the first row of which is just the set of roots of a different way) that this is a Galoisstyle group. To see this, we make the following observations: 1) Since is normal, hom . We claim (as did Galois, in where . 2) According to Theorem 2.8.3, for each , there is a that maps to . Furthermore, each element of determined by its value on . Hence, hom is uniquely Thus, letting arrangements as , we can rewrite the previous list of or Therefore, in the notation of Galoisstyle groups used earlier, so this list does indeed represent a Galoisstyle group. and Of course, Galois did not prove that his list (5.9.1) is a Galoisstyle group in the same way we have done. His first task is to show that each row of (5.9.1) is a permutation of the first row. 134 Field Theory The first step is to show that all of the elements of the table are roots of , that is, that for all and . For this, Galois considers the polynomials . Since it follows that the polynomial and the irreducible polynomial have a common root . Galois knew that this implies that . Hence, every root of is a root of , that is, for all and , as desired. Then Galois reasoned that if two elements where , are equal, then the polynomial above, and of the same row, has root and so, as which implies that all conjugates are roots of . In particular, . But these are roots from the first row of (5.9.1), which are distinct and so , a contradiction. For more details on Galois' approach to these issues, we refer to the reader to Edwards. Solvability by Radicals
So let us recap: Galois developed the notion of a Galois resolvent, that is, a primitive element of a splitting field of and showed that Galois resolvents always exist. He then used this notion to develop the concept of the Galoisstyle Galois group of . The stage is now set for his most famous result, namely, that the roots of a th or higher degree polynomial equation are not always solvable by radicals. Galois' approach was to consider the conditions imposed on the Galois group of a polynomial by the requirement that the polynomial equation be solvable by radicals. Here is a brief sketch. Note that since the roots of unity can be considered as known quantities (obtainable by the taking of roots), once a single root of a quantity is known, all other roots of that quantity, being of the form where is a root of unity, are also known. Since if , then it follows that an extension obtained by adjoining a single th root can be decomposed into a tower in which each step is obtained by adjoining a prime root of an element. Hence, a polynomial equation is solvable by Galois Theory I: An Historical Perspective 135 radicals if and only if a splitting field within a finite tower of fields for over can be "captured" (5.9.2) where each is a th root ( a prime) of some element in the previous field of the tower. Moreover, we may assume that the required roots of unity appear as necessary, in particular, we may assume that if is a step in the tower (5.9.2), then contains the th roots of unity. Now let us examine, as Galois did, the Galois groups from the definition that they form a nonincreasing sequence . It is clear (5.9.3) Moreover, if reverses inclusion, we have then, since the taking of Galois groups that is, . Galois studied the properties of the sequence (5.9.3). In particular, he showed that each group in (5.9.3) is a normal subgroup of its predecessor, and has prime index in its predecessor. A sequence of subgroups in which each group is normal in its immediate parent is called a normal series, and if the indices are prime, then the top group, which in Galois' case is , is called solvable. Galois proved that if is solvable by radicals, then its Galois group is solvable. He also proved the converse. Galois used his remarkable theory in his Memoir on the Conditions for Solvability of Equations by Radicals of 1831 (but not published until 1846), to show that the general equation of degree or larger is not solvable by radicals. It is worth noting that Ruffini, in 1799, offered the first "proof" that the th degree equation is not solvable by radicals. However, his proof was not completely convincing and a complete proof was given by Abel in 1826. Nevertheless, Galois' achievement is not diminished by these facts. 5.10 A Very Brief Look at the Life of Galois
Evariste Galois life was, to say the least, very short and very controversial. Of course, it would not be the subject of such legend today were it not for his remarkable discoveries, which spanned only a few short years. Galois was born on October 25, 1811, near Paris. Apparently, Galois was recognized at an early age as a brilliant student with some bizarre and rebellious tendencies. 136 Field Theory In 1828, at the age of 17, Galois attempted to enter the prestigious cole Polytechnique, but failed the entrance exams, so he remained at the royal school of LouisleGrand, where he studied advanced mathematics. His teacher urged Galois to publish his first paper, which appeared on April 1, 1829. After this, things started to go very badly for Galois. An article that Galois sent to the Academy of Sciences was given to Cauchy, who lost it. (Apparently, Cauchy had a tendency to lose papers; he had already lost a paper by Abel.) On April 2, 1829, Galois' father committed suicide. Galois once again tried to enter the cole Polytechnique, but again failed under some rather controversial circumstances. So he entered the cole Normale, considered to be on a much lower level than the cole Polytechnique. While at the cole Normale, Galois wrote up his research and entered it for the Grand Prize in Mathematics of the Academy of Sciences. The work was given to Fourier for consideration, who took it home, but promptly died, and the manuscript appears now to be lost. Galois possessed very strong political opinions. On July 14, 1831, he was arrested during a political demonstration, and condemned to six months in prison. In May 1832, Galois had a brief love affair with a young woman. He broke off the affair on May 14, and this appears to be the cause of a subsequent duel that proved fatal to Galois. Galois died on May 31, 1832. On September 4, 1843, Liouville announced to the Academy of Sciences that he had discovered, in the papers of Galois, the theorem, from his 1831 Memoir, that we mentioned earlier concerning the solvability by radicals of a prime degree equation, and referred to it with the words "as precise as it is deep." However, he waited until 1846 to publish Galois' work. In the 1850s, the complete texts of Galois' work became available to mathematicians, and it initiated a great deal of subsequent work by the likes of Betti, Kronecker, Dedekind, Cayley, Hermite, Jordan and others. Now it is time that we left the past, and pursued Galois' theory from a modern perspective. Chapter 6 Galois Theory II: The Theory 6.1 Galois Connections
The traditional Galois correspondence between intermediate fields of an extension and subgroups of the Galois group is one of the main themes of this book. We choose to approach this theme through a more general concept, however. Definition Let and be partially ordered sets. A Galois connection on the pair is a pair of maps and , where we write and , with the following properties: 1) (Orderreversing or antitone) For all and , and 2) (Extensive) For all , , and Closure Operations
Lurking within a Galois connection we find two closure operations. Definition Let be a partially ordered set. A map cl on (algebraic) closure operation if the following properties hold for all 1) (Extensive) cl 2) (Idempotent) cl cl cl is an : 138 Field Theory 3) (Isotone) cl An element elements in cl . The set of all closed is said to be closed if cl is denoted by Cl . be a Galois connection on and Theorem 6.1.1 Let . Then the maps are closure operations on cl . Moreover, 1) , that is, and , respectively, and we write cl and cl 2) , that is, cl Proof. Since cl cl , the orderreversing property of * gives and so , from which part 1) follows. Part 2) is similar. Theorem 6.1.2 The maps Cl and Cl are surjective and the restricted maps Cl Cl and Cl Cl are inverse bijections. Proof. Since cl , we see that is closed, that is, maps into Cl . Moreover, is surjective since if Cl , then cl . To see that is injective when restricted to closed elements, if Cl and , then , that is, . Similar arguments apply to . Finally, since cl we see that on Cl cl cl and similarly, cl on Cl . Theorem 6.1.3 Let be a Galois connection on a pair of lattices. 1) If is a complete lattice, then so is Cl , under the same meet as . A similar statement holds for . 2) De Morgan's Laws hold in Cl and Cl , that is, for Cl and Cl , and Galois Theory II: The Theory 139 Proof. For part 1), we apply Theorem 0.1.1 to the subset Cl of since has the property that cl , it follows that Suppose that . Then the meet exists in and since Cl all , we have cl whence cl holds and Cl lattice under meet in cl . First, Cl . for . Since the reverse inequality holds as well, equality . It follows from Theorem 0.1.1 that Cl is a complete . A similar argument can be made for . For part 2), observe first that and imply that and , whence . If and for Cl then and , whence . Thus, . It follows by definition of join that . The other parts of De Morgan's laws are proved similarly. Examples of Galois Connections
Our interest in Galois connections is the famous Galois correspondence between intermediate fields of a field extension and subgroups of the Galois group of an extension (to be defined later). However, let us take a look at some other examples of Galois connections. Example 6.1.1 Let and be nonempty sets and and be the corresponding power sets. Let be a relation on the maps for all and for all form a Galois connection on . . Then Example 6.1.2 Let and let be a field. Let be the set of all subsets of polynomials over in the variables . Let be the set of all subsets of , the set of all ordered tuples over . Let be defined by Set of all common roots of the polynomials in for all 140 Field Theory and let be defined by Set of all polynomials whose root set includes for all We leave it as an exercise to show that . is a Galois connection on Top and Bottom Elements
In many examples of Galois connections, elements. and have both top and bottom cl and similarly for . Note also A top element is closed, since that a top element is the image of the corresponding bottom element (if it exists), for is the image of and since , the image of must be at least as large as , and therefore equal to . However, a bottom element need not be closed. Indeed, the smallest closed element of is and so is closed if and only if , for example. In other words, a bottom element is closed if and only if it is the image of the corresponding top element. Indexed Galois Connections
Let denote the set of positive integers. In the set some obvious understandings about , in particular, for and implies , we observe for all , . Definition A Galois connection on is indexed if a) For each with , there exists a number , called the degree, or index of over . b) For each with , there exists a number , called the degree, or index of over . We generally write without a subscript to denote the appropriate index. Moreover, the following properties must hold: 1) (Degree is multiplicative) If or then 2) ( and are degreenonincreasing) If then If then Galois Theory II: The Theory 141 3) (Equality by degree) If or then If , then is said to be a finite extension of . If has a top and bottom element then the index of is index , and similarly for . From now on, when we write , it is with the tacit assumption that . The importance of indexing is described in the next theorem. It says that if a Galois connection is indexed, then the connection preserves the index of closed elements and that any finite extension of a closed element is also closed. Theorem 6.1.4 Let be an indexed Galois connection on . 1) (Degreepreserving on closed elements) If Cl and then . A similar statement holds for . 2) (Finite extensions of closed elements are closed) If Cl and then Cl . In particular, if is closed and is finite then all elements are closed. A similar statement holds for . Proof. For part 1), we have cl so equality holds throughout. For part 2), if Cl and cl and since is closed. , we may cancel to get cl then cl , which shows that cl Thus, in an indexed Galois connection, the maps are degreepreserving, orderreversing bijections between the collections of closed sets Cl and Cl . A Simple Degree Argument
There is a situation in which a simple degree argument can show that an element is closed. Referring to Figure 6.1.1, 142 Field Theory r cl(p) p p* r*
Figure 6.1.1 suppose that cl . Then is closed and since cl cl , we have Now, if then cl and so cl , that is, cl cl is closed. cl Theorem 6.1.5 If and one of the following holds 1) cl and 2) cl and then is closed. In particular, for , if then is closed. When is Closed The following nonstandard definition will come in handy. Definition For a Galois connection on , we say that is completely closed if every element of is closed, and similarly for . Also, the pair (or the connection) is completely closed if all elements of and all elements of are closed. We have remarked that the top elements and , if they exist, are always closed, but the bottom elements and need not be closed. However, the most important example of a Galois connection, namely, the Galois correspondence of a field extension , which is the subject of our investigations, has the property that is closed. So let us assume that is closed and see what we can deduce. Galois Theory II: The Theory 143 Since index index it follows that if has finite index, then so does . Hence, if either or has finite index, then is completely closed. Finally, if has finite index and is also closed, then the connection is completely closed. is closed) Let Theorem 6.1.6 ( be a Galois connection on , where and have top and bottom elements. Assume that is closed. Then index Also, 1) If index 2) If index index is completely closed. is completely closed. or index , then and is closed, then 6.2 The Galois Correspondence
Now we describe the main theme of the rest of the book. Definition The Galois group of a field extension the group Aut of all automorphisms of over also called the Galois group of over Note that when is algebraic, Aut and when is normal, hom Let and let be the complete lattice of all intermediate fields of , ordered by set inclusion. Let be the complete lattice of all subgroups of the Galois group , ordered by set inclusion. We define two maps and by hom , denoted by . The group , is is and fix where fix is called the fixed field of for all . These are pictured in Figure 6.2.1. 144 Field Theory E L K F GF(E) GK(E) GL(E) GE(E) = { } Figure 6.2.1The Galois correspondence Theorem 6.2.1 Let . The pair of maps fix is a Galois connection on called the Galois correspondence of the extension . Proof. It is clear from the definitions that both maps are orderreversing, that is, and fix Also, any element of fix , that is, is fixed by every element of fix Finally, any fixes every element in fix
fix , that is, Since and corollary. are complete lattices, Theorem 6.1.3 provides the following Corollary 6.2.2 The set Cl of closed intermediate fields and the set Cl of closed subgroups of are complete lattices, where meet is intersection. In particular, the intersection of closed intermediate fields is closed and the intersection of closed subgroups is closed. Note that both partially ordered sets and are topped and bottomed (as are all complete lattices). The top of is and the bottom is . The top of is and the bottom of is the trivial subgroup . Also, the image of the top is and so the bottom of is closed. Hence, three out of the four extreme elements are closed. We will spend much time discussing the issue of the closedness of the bottom element . Galois Theory II: The Theory 145 The Plan
Now that we have established that the Galois correspondence is a Galois connection, our plan is as follows. First, we will show that the Galois correspondence is indexed, where is the degree of and is the index of the subgroup in the group . Then we will describe the closed intermediate fields and the closed subgroups. The next step is to describe the connection between intermediate normal extensions and normal subgroups of the Galois group. (They don't call splitting fields normal extensions for nothing.) Finally, we describe the Galois group of a lifting and a composite. The Galois Correspondence Is Indexed
We would like to show that the Galois correspondence of an extension is indexed, where is the degree of the extension and is the index of the subgroup in the group . We know that the degrees are multiplicative and that The next theorem shows that the map is degreenonincreasing. Recall that if is finite, then . When is infinite, this inequality still holds provided that we interpret it, not as an inequality of infinite cardinals, but simply as saying that or . Theorem 6.2.3 For the tower , we have as elements of . hom Proof. Consider the function that maps to its restriction . Then if and only if and hom agree on , that is, if and only if . Hence is constant on the cosets of in and so induces an injection on , whence im But as elements of Showing that fix Theorem 6.2.4 Let fix and let fix fix , we have hom . is a bit more difficult. . Then 146 Field Theory Proof. First, if assume that is infinite, then there is nothing to prove, so let us , that is, is a finite set. Thus, is a complete set of distinct coset representatives for , and we may assume that . into . Then , then is a Let denote the set of all functions from vector space over , where if and Moreover, since the functions basis for over , we have dim defined by form a Thus, we have two vector spaces: fix is a vector space over fix of dimension fix fix and is a vector space over of dimension . We wish to show that dim fix dim . To do this, we will show that if over fix , then the evaluation functions fix are linearly independent , defined by are linearly independent over First, we must show that then . (In fact, the converse also holds.) is a welldefined function from and so to . If which implies that welldefined. So assume that fix by reindexing if necessary, let , that is, . Hence, is are linearly independent over fix and, be a nontrivial linear combination over that is shortest among all nontrivial if linear combinations equal to . Thus, for all . Dividing by necessary, we may also assume that . Thus (6.2.1) Then applying this to gives for all . Since the 's are fixed by any element of , and any has Galois Theory II: The Theory 147 the form for some , we deduce that (6.2.2) for all . In particular, if then (6.2.3) which implies, owing to the independence of the the 's can lie in fix . Let us assume that for which . We can replace by in (6.2.2) to get 's over fix , that not all of . Hence, there is a fix Applying gives for all and so Finally, subtracting (6.2.1) from (6.2.3) gives whose first coefficient is nonzero. But this is shorter than (6.2.1), a contradiction that completes the proof. Thus, the Galois correspondence of an algebraic extension We can now summarize our results in a famous theorem. is indexed. Theorem 6.2.5 (Fundamental Theorem of Galois Theory Part 1: The correspondence) The Galois correspondence of an extension is an indexed Galois connection and the bottom group is closed. It follows that the restrictions of and to closed elements are orderreversing, degreepreserving inverse bijections as well as lattice antiisomorphisms, that is, if are closed intermediate fields and are closed subgroups, then and fix fix fix fix We should note that the joins in the previous theorem are joins in the corresponding lattices. Thus, for instance, is the smallest closed subgroup of containing all of the subgroups , and this need not be the smallest subgroup of containing these groups. 148 Field Theory As a result of the closedness of following. Corollary 6.2.7 Let , Theorem 6.1.6 gives the be the Galois correspondence of . Then Also, 1) If 2) If 3) If , then is completely closed. , then is completely closed. and is closed, then and are completely closed. 6.3 Who's Closed?
We turn our attention to the question of which intermediate fields of an extension and which subgroups of the Galois group are closed. We know on general principles that top elements are always closed. Thus, and are closed. Moreover, the bottom group is also closed. We also know that any finite extension of a closed element is closed. Now we require a definition. Definition A normal separable extension or simply Galois. is called a Galois extension, The next theorem follows from the relevant properties of normal and separable extensions. Theorem 6.3.1 1) (Full extension Galois implies upper step Galois) Let . If is Galois then the upper step is Galois. 2) (Closed under lifting) The class of Galois extensions is closed under lifting. 3) (Closed under arbitary composites and intersections) The class of Galois extensions is closed under arbitrary composites and intersections. Let be algebraic. We wish to show that an intermediate field if and only if the extension is Galois. First, suppose that is closed and let of is also closed and so is closed . Then the finite extension Let be a complete system of distinct coset representatives for Galois Theory II: The Theory 149 . Each element of gives a distinct value on , that is, a distinct root of min , for if , then , which is not possible for . Hence, the roots of min are , which are distinct and lie in . Thus, is separable and min splits in , implying that is a Galois extension. For the converse, suppose that is Galois. If cl fix , has minimal polynomial min , then can have no roots other than . For if is a root of in some extension, then there is an embedding over for which . But since , it follows that and so . Thus has only one distinct root. Since is separable, it must be linear, which implies that . Thus, cl and is closed. Let us summarize, with the help of Theorem 6.2.7. Theorem 6.3.2 (Fundamental Theorem of Galois Theory Part 2: Who's closed?) Let be algebraic and consider the Galois correspondence on . 1) (Closed fields) The closed intermediate fields are precisely the fixed fields, that is, the fields of the form fix for some . a) An intermediate field is closed if and only if is Galois. b) Any extension of a closed intermediate field is closed. In particular, if is closed, then is completely closed. c) If cl and then is closed. In particular, if then is closed. 2) (Closed groups) The closed subgroups of are precisely the Galois groups of , that is, the subgroups of the form , for some intermediate field . a) Any finite extension of a closed subgroup is closed. b) is closed and so any finite subgroup of is closed. c) When is finite, so is and so is completely closed. 3) If is a finite Galois extension, then the correspondence is completely closed. As the next example shows, in the general algebraic case, not all subgroups need be closed. 150 Field Theory Example 6.3.1 For this example, we borrow from a later chapter the fact that for any prime power , there exists a finite field of size and if and only if . Referring to Figure 6.3.1, let and let . Since is a finite field, it is perfect and so is separable. Since is algebraically closed, . Hence is a Galois extension and therefore is closed. The extension is not finite, however, since and for all . E=Zp [E:P]>1 P GF(pq ) GF(pq) F=GF(p)=Zp
2 GF(E) (GF(E):H)>1? GP(E) H=< p> GE(E) = { }
Figure 6.3.1 Let be the subgroup of generated by the Frobenius map . The fixed field fix is the set of all for which , in other words, the roots in of the polynomial . But has roots in and so fix . It follows that cl
fix Hence, all we need do is show that closed. The key is that any has the form fixed set of is to conclude that is not for some and so the which is a finite set. Thus, we need only show that there is an element of that fixes infinitely many elements of . To this end, let be a prime and consider the field Then is a proper subfield of subfield . Hence is not trivial. But if , since it does not contain, for instance, the and since is Galois, the group , then fixes the infinite field . Galois Theory II: The Theory 151 Starting with a Field and a Subgroup of Aut The Galois correspondence begins with a field extension and the corresponding Galois group . Referring to Figure 6.3.2, we may also begin with a field and a subgroup of Aut . Then we can form the fixed field fix for all and consider the Galois correspondence of the extension fix , which we assume to be algebraic. The Galois group fix contains , but the containment may be proper. E Gfix(G)(E) G F=fix(G)
Figure 6.3.2Starting with a field {}
and a subgroup of Aut Since fix is algebraic and the base field fix is closed, it follows that fix is a Galois extension. Moreover, if fix , then the correspondence is completely closed (all intermediate fields and all subgroups are closed). We emphasize that may be a proper subgroup of its closure fix , as in Example 6.3.1. However, this does not happen if is finite, since finite subgroups are closed. Theorem 6.3.3 Let be a field and let be a group of automorphisms of . 1) If fix is algebraic, then it is Galois and all intermediate fields are closed. 2) If fix is finite, then all intemediate fields and all subgroups are closed. 3) If is closed (which happens if is finite), then is the top fix group of the correspondence. More on Closed Subgroups: Closure Points
Let be algebraic. The closure cl of a subgroup of the Galois group can be characterized in a useful way. The following nonstandard definition will help. 152 Field Theory Definition Let be algebraic. Let be a subgroup of the Galois group . A function is a closure point of if for any finite set , we have , that is, agrees with some member of on . Let denote the set of closure points of . First, note that a closure point of is a member of the Galois group in fact, is in the closure of , that is,
fix , cl Indeed, is a homomorphism because it agrees with a homomorphism on any finite set in and it fixes each element of fix because every member of fixes fix . We claim that cl . Since cl , the result would follow if were closed, but of course, it may not be. However, given any finite set , we need only work with the finite extension fix fix , whose fix Galois group is fix . In this case, all subgroups are closed. The problem is that we want to be in the Galois group and this requires that fix be normal. No problem really: we just pass to a normal closure. Consider the extension fix nc fix which is finite, normal, contains and has Galois group fix . Since all subgroups are closed, is a closed subgroup of the Galois group fix . Hence, in the Galois correspondence on fix cl It follows that any cl fix
fix fix , we have agrees with a member of on . But if , then
fix fix and so of on agrees with a member of on . Thus, cl , as desired. , that is, agrees with a member Theorem 6.3.4 Let be algebraic and let be a subgroup of the Galois group . Then cl is the set of closure points of . More specifically, the following are equivalent: 1) cl 2) For any finite set , we have . Galois Theory II: The Theory 153 Consequently, a subgroup of is closed if and only if it contains all of its closure points. In particular, any subgroup of the form contains all of its closure points. *The Krull Topology
For those familiar with elementary topology, we can make this discussion a bit more topological. We begin by extending the definition of closure point to apply to any set of functions in , not just subgroups of the Galois group. In particular, a function is a closure point of if for any finite set , we have . It is not hard to show that the operation is an algebraic closure operation, in the sense defined earlier in the chapter. In addition, we have and To see the latter, note that if , then for any finite subset , the function agrees with an element of on . But if , then there is a finite set for which does not agree with any element of on . Similarly, if , then there is a finite set for which does not agree with any element of on . However, is a finite set and so there must be some element that agrees with on , and therefore on both and , that is, on and on . But or , either one of which provides a contradiction. It follows that the operation is also a topological closure operation. Hence, the set of all complements of closed elements forms a toplology on . This topology is actually quite famous. Definition Let be the set of all functions from into . The finite topology on is defined by specifying as subbasis all sets of the form
, where , . Thus, a basis for consists of all sets of the form where , . To show that the topology obtained from closure points is the finite topology, let be any subset of . If is in the closure of under the finite topology, then any basis set that contains also contains an element of . It 154 Field Theory follows that for any finite set is, . In other words, , there is a for which is a closure point of . Thus , that . On the other hand, if , then agrees with some element of on any finite set and so any basis element containing must intersect , showing that . Thus, . Since the set of closed sets is the same in the topology of closure points and in the finite topology, these topologies are the same. Moreover, the Galois group is closed in the sense of closure points and so it is closed in the finite topology. Thus, the induced (subspace) topologies are the same and, in view of Theorem 6.3.4, we can state the following. Theorem 6.3.5 Let be algebraic. Then the Galois group is closed in the finite topology on . Moreover, a subgroup is closed in the Galois correspondence if and only if it is closed in the finite subspace topology on . The subspace topology of the finite topology inherited by is called the Krull topology on . We may phrase the previous theorem as follows: A subgroup of is Galoisclosed if and only if it is Krullclosed. Note that we do not say that the set of Galoisclosed subgroups of is the set of closed sets for a topology. We say only that these closed subgroups are that are Krullclosed in the Krull topology. There are other subsets of closed, for example, sets of the form which in general are not even groups. 6.4 Normal Subgroups and Normal Extensions
We now wish to discuss intermediate fields and their Galois groups . We begin with a result concerning the conjugates of a Galois group. Definition Let . If there is a then and are said to be conjugate. Theorem 6.4.1 1) If for which , , then for any hom , 2) If , then for any hom , Galois Theory II: The Theory 155 3) If with , then for any hom , 4) Let , with Galois. Then and are conjugate if and only if the Galois groups and are conjugate. Proof. For part 1), let . Then is an automorphism of . Moreover, since fixes , we have for , and so . Hence For the reverse inclusion, let automorphism of and if , then , where and so . Then is an which shows that . hom , then part 1) Part 2) follows from part 1), since when then any satisfies . Part 3) is similar. For part 4), if implies that Conversely, if and taking field fields gives Now, is normal in then part 1) implies that . if and only if for all . According to the previous theorem, and so if and only if If , then and so then taking fixed fields gives cl cl for all hom , that is, . For the converse, if Thus, if is closed, then , then for all and if, in addition, is normal. 156 Field Theory Note that when hom is defined by normal, the restriction map is a homomorphism, whose kernel is none other than the normal subgroup . Hence, the first isomorphism theorem of group theory shows that Moreover, if is normal, then is surjective, since any can be extended to an embedding of into over , which must be an element of . Hence, if , then Now we are ready to summarize. Theorem 6.4.2 (Fundamental Theorem of Galois Theory Part 3: Normality) hom Let . Let be the restriction map 1) If then and induces an embedding which is an isomorphism if the full extension is normal. 2) If and in addition, and is closed (that is, is Galois), then and induces an isomorphism 3) If is Galois, then if and only if . An Example
Now that we have a complete picture of the Galois correspondence, let us consider a simple example: the Galois correspondence of a splitting field for the polynomial over . Of course, is finite and Galois. Hence, the Galois correspondence is completely closed. The roots of this polynomial are and and so any member of the Galois group is a permutation of these roots. As to degree, we have Galois Theory II: The Theory 157 where the lower step has degree since min , by Eisenstein's criterion. The upper step has degree at most , but cannot be because , which does not contain . Hence, the upper step has degree and . One way to help find the Galois group is to look for an intermediate field that is normal, because the elements of are precisely the restrictions of the members of . Since any extension of degree is normal, we have . The elements of are the identity and the map . Since automorphisms of the roots of 1) 2) 3) 4) and 5) 6) 7) 8) which constitute the elements of . , we have min and can be extended to an element of . This gives ( ) and so each of the by sending to any Could be cyclic? Of course, one can tell this simply by checking for an element of order . A more elegant way is the following: If were cyclic, then all of its subgroups would be normal and so all of the intermediate fields would be normal extensions of . But is not normal, since does not contain all of the roots of min . Thus, all nonidentity elements have order or , and this is determined by whether or not . In particular, and all have order and and have order . Thus, has a normal cyclic subgroup where , and is the dihedral group of symmetries of the square. All nontrivial subgroups of have order correspond to the elements of order : 1) 2) 3) or . The subgroups of order 158 Field Theory 4) 5) The subgroups of order are the cyclic subgroup isomorphic to . A computation shows that 6) 7) 8) The lattice of subgroups is shown in Figure 6.4.1. and the subgroups G { , 2, 1, 2} { , 2, 3, 4} { , 2, 3, 4} { , 1} { , 2} { , 2} {}
Figure 6.4.1 Of course, the lattice of intermediate (fixed) fields is a reflection of this. To compute fixed fields, we use the fact that is a basis for over and is a basis for over and so the products form a basis for over . Hence, each has the form { , 3} { , 4} Thus for instance, fix if and only if , that is, Equating coefficients of the basis vectors gives fix As another example, note that fix fixes both and for all . Thus, and so (see Example 3.4.1). Moreover, is a normal subgroup of and so is a normal extension of degree . In fact, the roots of the polynomial are and and so is a splitting field for this polynomial. Galois Theory II: The Theory 159 More generally, the normal subgroups of correspond to the normal extensions of . These subgroups are , , the subgroups of order (index ) and . 6.5 More on Galois Groups
We now examine the behavior of Galois groups under lifting and under composites. As usual, we assume that all composites mentioned are defined. The Galois Group of a Lifting
Let be normal and let . Any , the Galois group of the lifting, is uniquely determined by what it does to (since it fixes ) and so the restriction map is an injection. Since is normal, it follows that . But may fix more than : It also fixes every element of that is fixed by , that is,
fix cl Note also that the restriction map is a homomorphism, and hence an embedding of into . We will show that this embedding is actually an cl isomorphism and
cl Note that if is Galois, then which simplifies the preceding to is Galois and so is closed, Theorem 6.5.1 (The Galois group of a lifting) Let . The restriction map
cl be normal and let where cl fix , defined by
cl is an isomorphism and Proof. We have already proved that is an embedding. It remains to show that is surjective. To avoid confusion, let us use the notation fix for the fixed field with respect to the Galois correspondence on , and fix for the fixed field with respect to the Galois correspondence on . Then fix im im for all for all for all fix Now, if we show that im is a closed subgroup with respect to the Galois 160 Field Theory correspondence on , it follows by taking Galois groups (of im
fix ) that and thus is surjective, completing the proof. If is finite, then all subgroups of the Galois group are closed, and we are finished. When is not finite, we must work a bit harder. We show that im is closed by showing that contains all of its closure points. So suppose that . To show that , we must find a for which . But any in is completely determined by its action on and so this completely determines , that is, if it exists. To this end, note that every has the form where and . Define a function by To see that is welldefined, let where agrees with and . Then since , there exists a on the elements , and so that Thus, is welldefined. Clearly, fixes and agrees with on . is Next, we show that is a closure point of . Then, since closed, it will follow that , and the proof will be complete. First note that on . Since , it agrees with some element of on any finite set . Hence, agrees with some element of on any finite subset of on any finite subset of . But , and so also with some element of also fixes and so agrees with any element of on . Thus, agrees with some element of on any finite subset of . But any finite subset of has the form , where element of is a finite subset of on . Thus, and so agrees with some , as desired. For a Galois extension , the previous theorem simplifies a bit. Galois Theory II: The Theory 161 Corollary 6.5.2 (The Galois group of a lifting) The lifting of a Galois extension by an arbitrary extension is Galois. Moreover, the restriction map defined by is an isomorphism and Also, 1) implies . 2) If is finite, then implies . Proof. We have proved all but the last two statements. Statement 1) is clear. As and the result to statement 2), since all is finite, we have follows by taking fixed fields. Corollary 6.5.2 yields a plethora of useful statements about degrees, all of which can be read from Figure 6.5.1. We leave details of the proof to the reader. EK E
finite Galois K E K F
Figure 6.5.1 Corollary 6.5.3 Suppose that is finite Galois and . Then 1) and so . If is also finite then 2) . 3) divides , with equality if and only if More generally, if is finite Galois for and finite then. letting when , we have 4) 5) if and only if for all . . is The Galois Group of a Composite
We now turn to the Galois group of a composite. Let and any is completely determined by its action on and by its restrictions and , or put another way, by the element . Then , that is, 162 Field Theory Indeed, the map is an embedding of groups. Moreover, as we will see, in the finite case, if the fields enjoy a form of independence ( ), then the embedding is an isomorphism. The following theorem gives the general case. Theorem 6.5.4 (The Galois group of a composite) 1) Let be a family of fields, with normal for all . Let be the direct product of the Galois groups and let be projection onto the th coordinate. Then the map defined by is an embedding of groups. Hence, is isomorphic to a subgroup of . 2) If is a finite family of finite Galois extensions, then the map is surjective and if and only if for all Proof. Since . , Theorem 6.4.1 implies that each individual restriction map is a surjective homomorphism from . Hence, is a homomorphism from As to the kernel of , if , then onto into , with kernel . and so on each is an embedding. , which implies that . Hence, ker and When is a finite family of finite Galois extensions, all Galois groups are finite and all subgroups and intermediate fields are closed. Since is injective, we have Galois Theory II: The Theory 163 im and also Hence is surjective if and only if gives the desired result. If and Corollary 6.5.3 is a finite Galois extension whose Galois group is a direct product , then we may wish to find intermediate fields whose Galois groups (over ) are isomorphic to the individual factors in the direct product. Corollary 6.5.5 Suppose that the form is a Galois extension with Galois group of If where is in the th coordinate and if fix then 1) is Galois, with Galois group 2) . 3) for all . Proof. Since , fix is closed and from Theorem 6.4.2 that and . is normal, it follows In addition, is Galois and since taking fixed fields gives . Hence, and Theorem 6.5.4 implies that for all . 164 Field Theory The Galois Group of the Normal Closure
We next wish to consider the Galois group of a normal closure, which is a special composite of fields. Theorem 6.5.6 Let 1) If be separable. nc then nc is isomorphic to a subgroup of hom hom 2) If, in addition to the conditions of part 1), is finite, then the direct product given above is a finite direct product. Proof. Let nc , the join being over all hom . Then hom Since is Galois, so is and Theorem 6.5.4 implies that is isomorphic to a subgroup of . The rest of part 1) follows from Theorem 6.4.1. For the second statement, if is finite, then hom and so the direct sum is a finite sum. 6.6 Abelian and Cyclic Extensions
Extensions are often named after their Galois groups. Here is a very important example. Definition A Galois extension is abelian if its Galois group abelian and cyclic if the Galois group is cyclic. is The basic properties of abelian and cyclic extensions are given in the next theorem, whose proof is left as an exercise. Note that abelian and cyclic extensions are not (quite) distinguished. Theorem 6.6.1 1) (Composite of abelian is abelian) If are abelian, then is abelian. 2) (Lifting of abelian/cyclic is abelian/cyclic) If is abelian (cyclic) and , then is abelian (cyclic). Galois Theory II: The Theory 165 3) (Steps in an abelian/cyclic tower are abelian/cyclic) If with abelian (cyclic), then and are abelian (cyclic). Abelian and cyclic extensions fail to be distinguished because, and only because if the steps in a tower are abelian (cyclic), this does not imply that the full extension is abelian (cyclic). What does it imply? Suppose that is a tower in which each step groups gives the series is abelian (cyclic). Taking Galois Consider the subtower follows from Theorem 6.4.2 that and that . Since the lower step is normal, it is a normal subgroup of its parent Since the latter is abelian (cyclic), so is the former. Thus, where each quotient group is abelian (cyclic). In the language of group theory, this series of subgroups is an abelian series. (When the groups are finite, the cyclic case and the abelian case are equivalent.) A group that has an abelian series is said to be solvable. Theorem 6.6.2 If is a tower of fields in which each step group is solvable. is abelian, then the Galois *6.7 Linear Disjointness
If and are finite extensions, the degree provides a certain measure of the "independence" of the extensions. Assuming that , we have The "least" amount of independence occurs when , or equivalently, when and the "greatest" amount of independence occurs when 166 Field Theory (6.7.1) We have seen (Corollary 6.5.3) that if one of the extensions is Galois, then (6.7.1) holds if and only if . For finite extensions in general, we cannot make such a simple statement. However, we can express (6.7.1) in a variety of useful ways. For instance, we will show that (6.7.1) holds for arbitrary finite extensions if and only if whenever is linearly independent over and is independent over then is also independent over . To explore the situation more fully (and for not necessarily finite extensions), it is convenient to employ tensor products. (All that is needed about tensor products is contained in Chapter 0.) The multiplication map defined by is bilinear and so there exists a unique linear map for which . Note that the image of form is the algebra of all elements of the for and algebraic, then . Hence, if or is algebraic, say and so the map is surjective. is If is a field, we use the term over . independent to mean linearly independent Theorem 6.7.1 Let and suppose that and are intermediate fields. Then and are linearly disjoint over if any of the following equivalent conditions holds. 1) The multiplication map is injective. 2) If is independent, then it is also independent. 3) If and are both independent, then is also independent. 4) If is a basis for over and is a basis for over , then is a basis for over . 5) There is a basis for over that is independent. Moreover, 6) and are linearly disjoint if and only if and are linearly disjoint, for all finite extensions and . 7) If and are linearly disjoint then Proof. 1 2 Let be independent and suppose that for . Since is injective and Galois Theory II: The Theory 167 we have Theorem 0.9.2 now implies that 2 3 Let and be for all . independent. If
, with then since equal , that is, is also independent, the coefficients of must for all . Since the 3 's are also independent, we get spans over for all , . and spans 4 This follows from the fact that if over then spans over . 1 The map sends a basis and is therefore injective. 4 for to a basis for Thus, each of 1) to 4) is equivalent, and by symmetry we may add the equivalent statement that any independent subset of is also independent. It is clear that 2) implies 5). 5 1 Let be a basis for over that is independent. Let basis for over . Then is a basis for over , for if
, be a with then since is independent, we have for all . Since the takes the basis 's are also independent, for all , . Finally, to the basis and so is injective. As to 6), it is clear that multiplication is injective if and only if each map is injective. Alternatively, if and are linearly disjoint, then so are and , for if is  168 Field Theory independent, then it is independent and hence also independent. Conversely, if were independent but failed to be independent, then some finite subset would be dependent as well, say , not all . Let and . Since is independent, it must also be independent by the linear disjointness of and . Thus, for all , a contradiction. For 7), suppose that and are linearly disjoint and . Then we have and where is a finite intermediate field in each case. It follows from part 6) that is linearly disjoint with itself. Therefore, if is a basis for over , it is also a basis for over and so , that is, and . Thus, . Corollary 6.7.2 (Linear disjointness in the finite case) Let suppose that and are intermediate fields of finite degree over . 1) and are linearly disjoint if and only if and for 2) If one of and only if or is Galois, then and are linearly disjoint if Proof. For part 1), if and are linearly disjoint, then part 4) of Theorem 6.7.1 implies that the degree condition above holds. Conversely, if this degree condition holds, and if is a basis for over and is a basis for over , then since the set spans and has size , it must also be a basis for . Hence, and are linearly disjoint. Alternatively, we have remarked that the multiplication map is surjective and so it is injective if and only if dim dim by Corollary 0.9.5 is equivalent to dim dim dim , which Part 2) follows from part 1) and Corollary 6.5.3. Exercises
1. 2. Find the Galois group of the polynomial over . Find the subgroups and intermediate fields. Prove that a pair of orderreversing maps between partially ordered sets is a Galois connection if and only if Galois Theory II: The Theory 169 3. 4. 5. 6. 7. 8. for all and , where and . Let . Prove that . fix If is an orderreversing bijection between two lattices, verify that and . Hint: first show that is also orderreversing. If and where is algebraic and and are Galois. Show that is a Galois extension. If is abelian, show that for every intermediate field we have . Let and be Galois extensions. Let be the join of and in the lattice of all subgroups of and let be the join in the lattice of all closed subgroups of . Show that is finite if and only if is finite, in which case . Let be finite. Let . Show that
fix fix 9. Let and let and be intermediate fields with and . Show that need not have degree a power of . Hint: The group has subgroups and . Consider the generic polynomial where are independent variables over . 10. Find an example of an infinite algebraic extension whose Galois group is finite. 11. Let be independent transcendentals over and consider the generic polynomial has coefficients . Then is algebraic over and so is algebraic. Show that the extension is Galois. Show that the degree of the extension is at most . Show that the Galois group of this extension is isomorphic to the symmetric group . 12. Prove Corollary 6.5.3. 13. Let be finite and Galois. Let be a prime for which , with . Show that for any , there is an intermediate field for which . Suppose that 170 Field Theory 14. Let be a perfect field. Define the order of a positive integer to be the largest exponent for which . Suppose that is a finite extension and that is a prime. Suppose that has order . Show that for any , has an extension whose degree has order . Show also that if is not a power of , then is not a power of . 15. Let be a finite Galois extension and let . Then divides . Use the following to show that the assumption that be Galois is essential. Let be the real cube root of , let be a cube root of . Let , and . 16. Prove the following statements about abelian and cyclic extensions. 1) If and are abelian, then is abelian. 2) If is abelian (cyclic) and , then the lifting is abelian (cyclic). 3) If with abelian (cyclic), then and are abelian (cyclic). 17. Let with roots in . Let . We can consider the splitting field of over as well as the splitting field of over . Note that . Let us examine the Galois groups and . a) If , show that , where fix fix b) Let be defined by . Show that is an isomorphism. 18. Referring to Theorem 6.5.4, show that if is an arbitrary family then the map defined by is an isomorphism if for all 19. Prove that is a topological group under the Krull topology. Show that is totally disconnected. 20. a) Show that in every Galois extension , there is a largest abelian ab ab subextension ab , that is, , is abelian and if ab with abelian then . b) If is a group, the subgroup generated by all commutators , for , is called the Galois Theory II: The Theory 171 21. 22. 23. 24. commutator subgroup. Show that is the smallest subgroup of for which is abelian. c) Let . If the commutator subgroup of a Galois group is closed, that is, if for some , ab then . Let . Show that the separable closure sc of in and the purely inseparable closure ic of in are linearly disjoint over . Moreover, if and if and ic are linearly disjoint over then is separable. Let and suppose that is a set of elements that are algebraically independent over . Then and are linearly disjoint over . Let and let . Assume that and are contained in a larger field. Then and are linearly disjoint over if and only if and are linearly disjoint over and and are linearly disjoint over . The following concept is analogous to, but weaker than, that of linear disjointness. Let and be extensions, with and contained in a larger field. We say that is free from over if whenever is a finite set of algebraically independent elements over , then is also algebraically independent over . a) The definition given above is not symmetric, but the concept is. In particular, show that if is free from over , then . Let be a finite algebraically independent set of elements of . Show that is algebraically independent over . b) Let and be field extensions, contained in a larger field. Prove that if and are linearly disjoint over , then they are also free over . c) Find an example showing that the converse of part b) does not hold. Chapter 7 Galois Theory III: The Galois Group of a Polynomial In this chapter, we pass from the highly theoretical material of the previous chapter to the somewhat more concrete, where we apply the results of the previous chapter to some special Galois correspondences. 7.1 The Galois Group of a Polynomial
The Galois group of a polynomial defined to be the Galois group of a splitting field , denoted by for over . If , is is a factorization of into powers of distinct irreducible polynomials over then is also a splitting field for the polynomial . , Moreover, the extension is separable (and hence Galois) if and only if each is a separable polynomial. To see this, let be the splitting field for satisfying . Then if is separable, so is the lower step and therefore so is . Conversely, if each factor is separable over , then is separably generated over and so is separable. Note that each is uniquely determined by its action on the roots of , since these roots generate , and this action is a permutation of the roots. In fact, if and are roots of , then there is a that sends to . Hence, the Galois group acts transitively on the roots of . However, not all permutations of the roots of need correspond to an element of . Of course, must send a root of an irreducible factor of to another root of the same irreducible factor, but even if is itself irreducible, not all permutations of the roots of correspond to elements of the Galois group. Thus, the Galois group is isomorphic to a transitive subgroup of the symmetric group , where . deg 174 Field Theory Let over and and and where deg and let be the splitting field for the splitting field for over . We clearly have and so Theorem 6.4.2 implies that or, in another notation, Theorem 7.1.1 Let where deg . The Galois group of is isomorphic to a quotient group of the Galois group of where is a splitting field for . 7.2 Symmetric Polynomials
In this section, we discuss the relationship between the roots of a polynomial and its coefficients. It is well known that the constant coefficient of a polynomial is the product of its roots and the linear term of is the negative of the sum of the roots. We wish to expand considerably on these statements. The Generic Polynomial and Elementary Symmetric Functions
If is a field and polynomial are algebraically independent over , the is referred to as a generic polynomial over of degree . Since the roots of the generic polynomial are algebraically independent, this polynomial is, in some sense, the most general polynomial of degree . Accordingly, it should (and does) have the most general Galois group , as we will see. It can be shown by induction that the generic polynomial can be written in the form Galois Theory III: The Galois Group of a Polynomial 175 where the coefficients are given by and are called the elementary symmetric polynomials in the variables . As an example of what can be gleaned from the generic polynomial, we deduce immediately the following lemma. Lemma 7.2.1 Let . The coefficients of elementary symmetric polynomials of the roots of are, except for sign, the . In particular, if has roots in a splitting field, then Since the extension is algebraic, the elementary symmetric polynomials are also algebraically independent over , . that is, there is no nonzero polynomial over satisfied by Theorem 7.2.2 The elementary symmetric polynomials are algebraically independent over . Proof. Since , where the upper step is algebraic, Theorem 4.3.2 implies that contains a transcendence basis for over . But is a transcendence basis and so : . Hence, is a transcendence basis. The Galois Group of the Generic Polynomial
Let us compute the Galois group of is a splitting field for over no multiple roots, the extension over , and since . Since has is finite and Galois and so We claim that is isomorphic to the symmetric group , define a map . Let . For any by Since the 's are algebraically independent over , this is a welldefined automorphism of over , which fixes the elementary symmetric 176 Field Theory polynomials . Thus, , that is, is an automorphism of , where over for any Moreover, each is distinct, since if . It follows that is isomorphic to . , then and for all and so Theorem 7.2.3 Let be algebraically independent over and let be the elementary symmetric polynomials in . 1) The extension is Galois of degree , with Galois group isomorphic to the symmetric group . 2) fix , that is, any rational function in that is fixed by the maps is a rational function in . 3) The generic polynomial is irreducible over . Proof. To prove part 3), observe that if were equal to where deg and deg , then the Galois group of would have size at most . Hence is irreducible. Symmetric Polynomials
Now we are ready to define symmetric polynomials (and rational functions). Definition A rational function if is symmetric in for all permutations , that is, if the Galois group of the extension fix . , where is A famous theorem of Isaac Newton describes the symmetric polynomials. Theorem 7.2.4 (Newton's Theorem) Let be algebraically independent over and let be the elementary symmetric polynomials in . 1) A polynomial is symmetric in if and only if it is a polynomial in , that is, if and only if for some polynomial over . Moreover, if has integer coefficients, then so does . 2) Let . Then the set of symmetric polynomials over in the roots of is equal to the set of polynomials over in the coefficients of . Galois Theory III: The Galois Group of a Polynomial 177 In particular, any symmetric polynomial over in the roots of is an element of . 3) Let be a polynomial with integer coefficients. Then the set of symmetric polynomials over in the roots of is equal to the set of polynomials over in the coefficients of . In particular, any symmetric polynomial over in the roots of is an integer. Proof. Statements 2) and 3) follow from statement 1) and Lemma 7.2.1. If has the form , then it is clearly symmetric. For the converse, the proof consists of a procedure that can be used to construct the polynomial . Unfortunately, while the procedure is quite straightforward, it is recursive in nature and not at all practical. We use induction on . The theorem is true for , since . Assume that the theorem is true for any number of variables less than and let be symmetric. By collecting powers of , we can write where each is a polynomial in . Since is symmetric in and are independent, each of the coefficients is symmetric in . By the inductive hypothesis, we may express each as a polynomial in the elementary symmetric polynomials on . If these elementary symmetric polynomials are denoted by , then (7.2.1) where each is a polynomial in integer coefficients. Note that the symmetric functions functions as follows , with integer coefficients if has can be expressed in terms of the symmetric (7.2.2) These expressions can be solved for the 's in terms of the 's, giving 3 3 3 3 and from the last equation in (7.2.2), 178 Field Theory (7.2.3) Substituting these expressions for the 's into (7.2.1) gives where each is a polynomial in and , with integer coefficients if has integer coefficients. Again, we may gather together powers of , to get where each is a polynomial in , with integer coefficients if has integer coefficients. If , we may reduce the degree in by using (7.2.3), which also introduces the term . Hence, (7.2.4) where each is a polynomial in integer coefficients. , with integer coefficients if has Since the left side of (7.2.4) is symmetric in the 's, we may interchange , for each , to get and valid for all . Hence, the polynomial has degree (in ) at most but has must be the zero polynomial. Thus, , as desired. Example 7.2.1 Let roots in a splitting field. For distinct roots for and , whence it be a polynomial with , the polynomials are symmetric in the roots of , and so Theorem 7.2.4 implies that the 's can be expressed as polynomials in the elementary symmetric polynomials of the roots. One way to derive an expression relating the 's to the 's is by following the proof of Theorem 7.2.4. In the exercises, we ask the reader to take another approach to obtain the socalled Newton identities for of the . These identities can be used to compute recursively the 's. 's in terms Galois Theory III: The Galois Group of a Polynomial 179 7.3 The Fundamental Theorem of Algebra
The Galois correspondence can be used to provide a simple proof of the fundamental theorem of algebra. As an aside, the history of the fundamental theorem is quite interesting. It seems that attempts to prove the fundamental theorem began with d'Alembert in 1746, based on geometric properties of the complex numbers and the concept of continuity, which was not well understood at that time. In 1799, Gauss gave a critique of the existing "proofs" of the fundamental theorem, showing that they had serious flaws, and attempted to produce a rigorous proof. However, his proof also had gaps, since he suffered from the aforementioned lack of complete understanding of continuity. Subsequently, in 1816, Gauss gave a second proof that minimized the use of continuity, assuming a form of the intermediate value theorem. It was not until Weierstrass put the basic properties of continuity on a rigorous foundation, in about 1874, that d'Alembert's proof and the second proof of Gauss could be made completely rigorous. We will also assume a form of the intermediate value theorem, namely, that if is a real polynomial, and if and have opposite signs, for , then there is a for which . From this, one can deduce that any odd degree real polynomial must have a real root and is therefore reducible over . It follows that any nontrivial finite extension of must have even degree, since it must contain an element whose minimal polynomial has even degree. We also require some knowledge of complex numbers, namely, that every complex number has a complex square root, which can be seen from a geometric perspective: implies . Hence, no complex quadratic is irreducible over , since the method of completing the square shows that the roots of lie in . It follows that has no extensions of degree . Theorem 7.3.1 (The fundamental theorem of algebra) Any nonconstant polynomial over has a root in , that is, is algebraically closed. Proof. We first show that it is sufficient to prove the theorem for real polynomials. Let be nonconstant. Consider the polynomial , where the overbar denotes complex conjugation of the coefficients. Then is a real polynomial and has a complex root if and only if has a complex root. Hence, we may assume that . Now consider the tower over . Since , where is a splitting field for divides , we conclude 180 Field Theory that , for some with , showing that splits over . be a Sylow subgroup of fix odd. Our goal is to show that Let . Then and so Since has no nontrivial extensions of odd degree, we deduce that is a group of order . and Thus, we have the tower in which . Therefore, according to Theorem 0.2.19, has a subgroup of any order dividing . But cannot have a subgroup of order that is, index because then fix fix fix and so , which implies which is not possible. Hence, that , whence . 7.4 The Discriminant of a Polynomial
We have seen that the Galois group of a polynomial of degree is isomorphic to a subgroup of the symmetric group and that the Galois group of a generic polynomial is isomorphic to itself. A special symmetric function of the roots of , known as the discriminant, provides a tool for determining whether the Galois group is isomorphic to a subgroup of the alternating group . Let be a polynomial over , with roots in a splitting field E. Let The discriminant of is Note that if and only if , which is clearly symmetric in the roots. has no multiple roots. Let us assume that . Then is the product of distinct separable polynomials, implying that is a Galois extension. Hence, fix Since for all theorem also implies that , we deduce that .) . (Newton's Galois Theory III: The Galois Group of a Polynomial 181 Each transposition of the roots sends to , and so for any , where is if is an even permutation and if is an odd permutation. Thus, the location of can give us some information about the parity of the permutations in the Galois group. If char , then for all and so is always in the base field . This is not very helpful. But if char , then fixes if and only if is an even permutation. Put another way, if and only if contains only even fix permutations, that is, . If then must contain an odd permutation. It is not hard to show that if a subgroup of contains an odd permutation then the subgroup has even order and exactly half of its elements are even. Hence, if is, then has even order and , that Since all groups are closed, it follows that fix Since and fix fix Thus, is the fixed field of the subgroup of even permutations in Let us summarize. Theorem 7.4.1 Let have degree and splitting field . Let be any square root of the discriminat of . 1) if and only if has multiple roots in . 2) Assume that and char . a) if and only if is isomorphic to a subgroup of . b) if and only if is isomorphic to a subgroup of that contains half odd and half even permutations. In this case, fix 3) If and char , then isomorphic to a subgroup of . but need not be . , we have 182 Field Theory Proof. For part 3), recall that the generic polynomial has Galois group over . The usefulness of Theorem 7.4.1 comes from the fact that can actually be computed without knowing the roots of explicitly. This follows from the fact that is the Vandermonde determinant Multiplying this by its transpose gives where . Newton's identities can then be used to determine the 's in terms of the coefficients of the polynomial in question (see Example 7.2.1 and the exercises). We will see some examples of this in the next section. 7.5 The Galois Groups of Some SmallDegree Polynomials
We now examine the Galois groups of some smalldegree polynomials. The Quadratic
Quadratic extensions (extensions of degree ) hold no surprises. Let be a quadratic over observe that , with splitting field and . To compute the discriminant, Hence a familiar quantity. Multiple Roots If , then has a double root and Galois Theory III: The Galois Group of a Polynomial 183 The root will lie in for most wellbehaved base fields . In particular, if char , then implies . If char and is perfect a must be reducible over and so finite field, for example) then . However, the following familiar example shows that may have a multiple root not lying in . Let where is transcendental over and let Since root , this polynomial is irreducible over . , but has a multiple No Multiple Roots If , then has distinct roots and there are two possibilities: 1) The roots lie in , is reducible and is trivial. 2) The roots do not lie in , is irreducible and generated by the transposition of the roots. is Thus, when char , we can tell whether the roots lie in by looking at the discriminant: If , then and possibility 2) obtains. Of course, this is also evident from the quadratic formula If , then . Hence the roots lie in if and only if . (We can now rest assured that what we tell our children about quadratic equations is actually true.) Theorem 7.5.1 Let have degree . 1) If then has a double root . If char or is perfect, then . In any case, is trivial. 2) If then has distinct roots and there are two possibilities: a) The roots lie in , is reducible and is trivial. b) The roots do not lie in , is irreducible and is generated by the transposition of the roots. When char , we can distinguish the two cases as follows: Case 1) holds if and case 2) holds if . Let us turn now to a more interesting case. 184 Field Theory The Cubic
Let
3 have splitting field in . . Then is irreducible if and only if none of its roots lie If splits over then and its Galois group is trivial. If reducible but does not split, then it can be factored over : is where is irreducible over Galois group is isomorphic to . Now let us assume that . Hence, and the is irreducible. A lengthy computation gives If , then has multiple roots and since each root must have the same multiplicity, we are left with char and Hence, the extension Galois group is trivial. If is purely inseparable of degree and the , then has no multiple roots and is therefore separable. Hence, is Galois and which leaves the possibilities and . We can now give a complete analysis for the cubic. Note that when char , knowledge of irreducibility and the value of determine the Galois group and the splitting field. Theorem 7.5.2 (The cubic) Let have degree , with splitting field and Galois group . Then there are four mututally exclusive possibilities, each of which can be characterized in four equivalent ways: 1) a) b) is the splitting field for c) d) (For char ) is reducible and . 2) a) Galois Theory III: The Galois Group of a Polynomial 185 b) is reducible and a root not in . is a splitting field for , where is c) d) (For char ) is reducible and . 3) a) b) is irreducible and is the splitting field for , for any root c) d) (For char ) is irreducible and . 4) a) b) is irreducible and is the splitting field for , for any root c) d) (For char ) is irreducible and . Proof. We leave proof to the reader. We know that . For , we can learn more about the roots of a cubic by looking at the sign of . A cubic over has either one reat root and two nonreal roots or three real roots and . In the former case, and so . In the latter case, . Theorem 7.5.3 (The cubic over ) Let have degree . Then 1) if and only if has exactly one real root 2) if and only if has three real roots. Example 7.5.1 Let over . Any rational root of must be (Theorem 1.2.3) and so is irreducible. The discriminant is , so has three real roots. Since , we have and has splitting field , for any root . On the other hand, for any prime , the polynomial is irreducible over and has discriminant , whose square root is not in . Hence, has one real root and two nonreal roots, the Galois group of is isomorphic to and has splitting field . *The Quartic
Since the Galois group of an irreducible quartic polynomial is isomorphic to a transitive subgroup of , we should begin by determining all such subgroups. Theorem 0.3.2 implies that if is a transitive subgroup of then 186 Field Theory or Here is a list. 1) (Order : cyclic group) The cyclic group occurs as a subgroup of The elements of of order are the cycles The three subgroups of isomorphic to are . . 3 2) (Order : Klein fourgroup) The Klein fourgroup subgroup of . In particular, let , , , occurs as a which is isomorphic to . We leave it to the reader to show that is normal in . Note also that . This and the previous case exhaust all nonisomorphic groups of order . The group contains other isomorphic copies of the Klein four group, such as However, suppose that such a subgroup is transitive. Every nonidentity element has order and so is a product of disjoint cycles (transpositions). Hence, is a transposition or a product of two disjoint transpositions. But a transposition links only two elements of together and a product of disjoint transpositions links two pairs of elements together. Since there are pairs that must be linked, we deduce that contains no transpositions and therefore must be . 3) (Order : dihedral group) The dihedral group of symmetries of the square, thought of as permutations of the corners of the square, occurs as a subgroup of of order . These subgoups are Sylow subgroups , , , , , , Note that , for each . 4) (Order : alternating group) The alternating group is the only subgroup of of order . 5) (Order : symmetric group) Of course, is the only subgroup of of order . Galois Theory III: The Galois Group of a Polynomial 187 Now let be an irreducible quartic over and assume that char . This will insure that , that is separable and that all irreducible cubic polynomials that we may encounter are separable. Replacing by of the form will eliminate the cubic term, resulting in a polynomial which is often referred to as the reduced polynomial for . The polynomials and have the same splitting field and hence the same Galois group, and their sets of roots are easily computed, one from the other. Let be the splitting field of , let be the roots of in and let be its Galois group. For convenience, we identify with its isomorphic image in the permutation interchanges and . . For example, To analyze the quartic , we want to find a strategically placed intermediate field. One way to do this is to find a strategically placed subgroup of the Galois group, one that has nice intersection properties with the candidates listed above.. The alternating group immediately springs to mind, but this may be too large. In fact, if then is a subgroup of . So let us try the Klein four group , which gives us a subgroup of , as shown in Figure 7.5.1. E = split(p(x))
e =2 or 4 GF(p(x)) = G
d = 1: [r(x) splits over F] d = 2: [r(x) has one root in F] d = 3 or 6: [r(x) irred. over F] fix(V G) = split(r(x))
d = 1: [r(x) splits over F] d = 2: [r(x) has one root in F] d = 3 or 6: [r(x) irred. over F] V G
e =2 or 4 F
Figure 7.5.1 Comparing with the candidates for 1) 2) , we have 188 Field Theory 3) 3) 4) 5) 6) Thus, 1) 2) 3) , , for , , or or We next determine the fixed field the expressions fix . Each element of fixes and so . By checking each permutation in see that no permutation outside of fixes , and . Thus, , it is not hard to Taking fixed fields gives fix We would like to show that and so is the splitting field for the cubic polynomial over , but this requires that the coefficients of lie in . The coefficients of are the elementary symmetric polynomials of the roots and and since every permutes , and , it follows that any symmetric function of , and is fixed by and so lies in . Thus, is the splitting field for the cubic . Hence, or as shown in Figure 7.5.1. Definition The polynomial resolvent cubic of is called the . . First note that . Then if we The Coefficients of the Resolvent Cubic Now let us determine the coefficients of the resolvent cubic since has no cubic term, it follows that Galois Theory III: The Galois Group of a Polynomial 189 write then and quadratic polynomials over , say . Now write as a product of where the linear coefficients are negatives of each other since term, and where the roots of the first factor are and . Then has no cubic and so Multiplying out the expression for equations and equating coefficients gives the Solving the first two for and and substituting into the third gives and so satisfies the polynomial and satisfies the polynomial But we can repeat this arguement, factoring which the roots of the first quadratic are and into a product of quadratics for , say and so that and . Similarly, . The same algebra as before leads to the fact and so is the resolvent cubic of . Final Analysis of the Quartic The first thing to note is that the discriminants of and are equal: . We leave verification of this as an exercise. Let be the Galois group of and let be the Galois group of . The following can be gleaned from Theorem 7.5.2. 1) If 2) If is reducible and and so is reducible and ), then (in which case splits over ), then . Hence, . (in which case has a single root in and there are two possibilities. If 190 Field Theory then and so for or , or . But is not possible, so . Note that in this case, since is the splitting field for over and the polynomial must have an irreducible quadratic factor over . If then and , for or . In this case, is irreducible over . 3) If is irreducible and , then and . Hence , which implies that and so . Thus . 4) If is irreducible and , then and is not a subgroup of . Hence . and so or . But and so . Theorem 7.5.4 (The quartic) Let be an irreducible quartic over a field splitting field for over . Let , with char . Let be the be obtained from by substituting for and let be the resolvent cubic of . Let be the Galois group of and let be the Galois group of . Then . If is reducible over then 1) If , then . 2) If , there are two possibilities. Let fix . a) and , for or which occurs if and only if is reducible over , in which case has an irreducible quadratic factor over . In this case, . b) and , for or , which occurs if and only if is irreducible over . In this case, and . If is irreducible over then 3) If , then . 4) If , then . The Quartic Consider the special quartic and let Galois Theory III: The Galois Group of a Polynomial 191 If we denote the roots of and in by , in this order, then The roots and of are given by The square root of the discriminant of is and since base field . is invariant under each . Hence, if and only if , it must lie in the , or equivalently, Let us also note that instance, sends to is fixed by every possible choice of sends to and . It follows that fix . . For The irreducibility of over can be determined as follows. Certainly if is reducible over , then so is . On the other hand, if is irreducible then its roots and do not lie in , whence cannot have a linear factor over and, if so reducible, must have the form where, as seen by equating coefficients, . However, if then which gives contradicting the irreducibility of summarize as follows: 1 2 If If then then . Thus, and . We can , and therefore , is reducible. is reducible if and only if it has the form where and . 192 Field Theory For example, let From 2 , we have over and . Then , . and since the latter has no solutions in Let us now assume that irreducible and , and that The resolvent cubic for , we see that is irreducible over . is irreducible. It follows that is also . Recall also that if and only if fix . (which is already in reduced form) is which is definitely reducible. Hence, Theorem 7.5.4 tells us the following. 1) If 2) If a) , then . , there are two possibilities. Let fix . and , for or which occurs if and only if is reducible over , in which case has an irreducible quadratic factor over . In this case, . b) and , for or , which occurs if and only if is irreducible over . In this case, and . Case 1) above is straightforward. Referring to case 2), we have and . But in both cases, and so . Also, it appears that we could use some more information about when is irreducible over . Lemma 7.5.5 Assume that is reducible and . Then and 1) is irreducible over if and only if is irreducible over . 2) is irreducible over if and only if . Proof. For part 1), if is reducible over , then clearly reducible over . Conversely, suppose that is reducible over and is where assumption. Thus . If and , then and so , contrary to Galois Theory III: The Galois Group of a Polynomial 193 which implies that is reducible over over , then we can assume that . If has a linear factor and so which shows that is irreducible over . if and only if , we have Finally, it is clear that the quadratic is reducible over . But under the assumption that For if then squaring gives and since , we must have so , whence and so then . But since . and . Conversely, if We can now give a complete analysis for this quartic. Theorem 7.5.6 (The irreducible quartic ) Let be a quartic over a field , with char . Let be the splitting field for over and let be its Galois group. 1) If then splits over and . 2) If , then there are two possibilities: a) If , then has an irreducible quadratic factor and , for or . b) If , then , for or . Exercises
1. 2. Prove that part 4a) and part 4b) of Theorem 7.5.2 are equivalent. Let be a prime. Let be an irreducible polynomial of degree with exactly two nonreal roots. Prove that the Galois group of is . Hint: Recall that is generated by a cycle and a transposition. Use Cauchy's theorem on . What is the transposition? Let where are algebraically independent over . Show that is irreducible over , separable and its Galois group is isomorphic to . Thus, if the roots of are then are algebraically independent over if and only if are. 3. 194 Field Theory is a quartic polynomial with resolvent cubic then . 5. Find the Galois groups of the following polynomials over : a) b) c) 6. Suppose that is irreducible over and that is isomorphic to . What are the possible degrees of ? 7. Suppose that is irreducible of degree and let be a root of in . What are the possibilities for , expressed in terms of ? 8. If has roots then . 9. Let , where and are algebraically independent over . Let be the elementary symmetric polynomials on , and . Show that but the Galois group of over is isomorphic to . 10. Let 4. If be the generic polynomial with algebraically independent roots . Let . Since the 's are symmetric polynomials in the roots of , Theorem 7.2.4 implies that they can be expressed as symmetric polynomials in the elementary symmetric polynomials . Newton's identities are valid for , where this reduces to and for . Note that for , Prove these identities as follows: a) For , consider the sum . b) For , consider the sum . c) For , proceed by induction on . Let and write the coefficients of as . Then Newton's identites are Denote the left side of this by . Show that Galois Theory III: The Galois Group of a Polynomial 195 Hence, . Show that . Is this possible? d) Let . Find the values of and find the discriminant of . 11. This exercise concerns the issue of when a value that is expressed in terms of nested radicals where (char ) can be written in terms of at most two unnested radicals. For instance, we have but the number the quartic cannot be so written. Note that is a root of Assume that some and in is irreducible over if and only if . Show that for Chapter 8 A Field Extension as a Vector Space In this chapter, we take a closer look at a finite extension from the point of view that is a vector space over . It is clear, for instance, that any is a linear operator on over . However, there are many linear operators that are not field automorphisms. One of the most important is multiplication by a fixed element of , which we study next. 8.1 The Norm and the Trace
Let by be finite and let . The multiplication map is an linear operator on , since defined for all and . We wish to find a basis for which the matrix of has a nice form. over under Note that if , then for all and so as an element of if and only if is the zero operator on . Hence, the set of polynomials over satisfied by is precisely the same as the set of polynomials satisfied by . In particular, the minimal polynomial of in the sense of fields is the same as the minimal polynomial of the linear operator . The vector subspace over and if
, of . If is invariant under the linear operator , since is an ordered basis for then the matrix of ordered basis for products over with respect to where is , . If is an , then the sequence of 198 Field Theory is an ordered basis for note that over . To compute the matrix of with respect to , , and so each of the subspaces is also invariant under . Hence, the matrix of is also equal to , and the matrix of with respect to the ordered basis has the block diagonal form (8.1.1) It follows that if the characteristic polynomial of characteristic polynomial of is is , then the The wellknown CayleyHamilton theorem implies that therefore . But is monic and has degree deg min , whence min . Theorem 8.1.1 Let operator on defined by and be finite and let . If is the linear , then the characteristic polynomial of is min We recall from linear algebra that if is a linear operator on a finite dimensional vector space over , the trace of is the sum of the eigenvalues of and the norm (determinant) of is the product of the eigenvalues of , in both cases counting multiplicities. Recall also that (as with all symmetric polynomials in the roots of a polynomial) the trace and the norm lie in the base field . We are motivated to make the following definition. Definition Let be finite and let . The trace of over , denoted by Tr , is the trace of the linear operator and the norm of over , denoted by , is the norm of . Note that the trace and norm of on the element itself. depend on the extension field , and not just Since the trace of a linear operator is the sum of the roots of its characteristic polynomial and the norm is the product of these roots, Theorem 8.1.1 allows us A Field Extension as a Vector Space 199 to express the trace and norm in terms of the roots of the minimal polynomial of on the subfield . Let be finite, let and let min have roots Tr and in a splitting field. It follows from Theorem 8.1.1 that We remark that many authors simply define the trace and norm of from these formulas. In terms of distinct roots of has multiplicity (Theorem 3.5.1) and so Tr and directly , if these are , then each of these roots , where is the radical exponent of We can also express the trace and norm in terms of embeddings. Let hom where . If and min , then is a list of the roots of in . However, each distinct root appears times in this list, since this is the number of ways to extend an embedding of to an embedding of , and each such extension has the same value at . Hence, and 200 Field Theory These formulas will provide another expression for the norm and the trace. Let us summarize. Theorem 8.1.2 Let . 1) If has roots Tr and distinct roots then be finite and let with min and 2) If hom Tr and then if if is separable is inseparable Proof. As for the first statement in part 2), if is inseparable, then , char and , whence Tr . Theorem 8.1.2 can be used to derive some basic properties of the trace and the norm. Theorem 8.1.3 Let 1) The trace is an , Tr be finite. linear functional on Tr , that is, for all Tr and A Field Extension as a Vector Space 201 2) The norm is multiplicative, that is, for all , Also, for all , 3) If then Tr and then and 4) If Tr are finite and if Tr Tr Proof. We prove part 4), leaving the rest for the reader. Let and let hom and hom Extend each to an embedding of which is an embedding of into over hom Note that these embeddings are distinct, for if fixes and so , that is, Hence, . Moreover, since hom hom it follows that hom Now, for the norm statement, we have from Theorem 8.1.2, hom , then , which implies that . and consider the products , that is, , each 202 Field Theory Proof of the statement about the trace is similar. *8.2 Characterizing Bases
Let be finite and separable. Our goal in this section is to describe a condition that characterizes when a set of vectors in is a (vector space) basis for over . Bilinear Forms
In order to avoid breaking the continuity of the upcoming discussion, we begin with a few remarks about bilinear forms. For more details, see Roman, Advanced Linear Algebra. If is a vector space over , a mapping is called a bilinear form if it is a linear function of each coordinate, that is, if for all and , For convenience, if , we let A bilinear form is symmetric if for all together with a bilinear form is called a metric vector space. . A vector space Definition Let be a metric vector space. 1) A vector is degenerate if it is orthogonal to all vectors in (including itself), that is, if 2) The space is degenerate (or singular) if it contains a nonzero degenerate vector. Otherwise, it is nondegenerate (or nonsingular). 3) The space is totally degenerate (or totally singular) if every vector in is degenerate, that is, if the form is the zero function A Field Extension as a Vector Space 203 for all If with respect to . is an ordered basis for is over , the matrix of the form The proof of the following theorem is left to the reader. Theorem 8.2.1 1) Let be the matrix of a bilinear form on basis . If then , with respect to the ordered where is the coordinate matrix for with respect to . 2) Two matrices and represent the same bilinear forms on , with respect to possibly different bases, if and only if they are congruent, that is, if and only if for some invertible matrix . 3) A metric vector space is nonsingular (nondegenerate) if and only if any, and hence all, of the matrices that represent the form are nonsingular. Characterizing Bases
As mentioned earlier, for a finite separable extension describe a condition that characterizes when a set is a basis for over . Suppose that hom We will show that is nonsingular: is a basis for over if and only if the following matrix are vectors in , where , we wish to of vectors in and let Our plan is to express this matrix in terms of the matrix of a bilinear form. To this end, observe that for any vectors and in , 204 Field Theory , Tr and so Tr In particular, Tr We can now define a symmetric bilinear form on Tr This form has a rather special "all or nothing" property: If degenerate vector , then for any , we have by (8.2.1) contains a nonzero and so is totally degenerate. In other words, totally degenerate. is either nondegenerate or else We have assumed that the extension is finite and separable. Of course, if we drop the separability condition, then the matrix is no longer square and therefore cannot be invertible. However, the bilinear form (8.2.1) still makes sense. As it happens, this form is nonsingular precisely when is separable. Theorem 8.2.2 Let be finite. The following are equivalent: 1) is separable 2) is nondegenerate. When is separable, the matrix is nonsingular if and only if is a basis for over . Proof. If is inseparable, then part 2) of Theorem 8.1.2 shows that the trace is identically , whence is totally degenerate. Thus, if is nondegenerate, then is separable. For the converse, since is finite and separable, it is simple, that is, . If , then , is an ordered basis for over and A Field Extension as a Vector Space 205 Tr But is a Vandermonde matrix, for which it is well known that det Moreover, since each hom is uniquely determined by its value on the primitive element , the elements are distinct and so det . Hence det is also nonzero and is nondegenerate. For the final statement, suppose first that is nonsingular. If for , then applying gives where and is the th column of . Hence, and the nonsingularity of implies that , that is, for all . Hence, is linearly independent and therefore a basis for over . For the converse, if matrix of the form (8.2.1) is Tr and since is nonsingular because is also nonsingular. is nondegenerate, the matrix is an ordered basis for over , then the The Algebraic Independence of Embeddings
Let and be fields. Recall that the Dedekind independence theorem says that any set of distinct embeddings of into is linearly independent over . To put this another way, let and consider the linear polynomial Then the Dedekind independence theorem says that if is the zero 206 Field Theory map, then is the zero polynomial. Under certain circumstances, we can strengthen this result by removing the requirement that be linear. Let be finite and separable of degree hom If function from is a polynomial with coefficients in into , defined by , then is a and let For example, if then (Note that we are not composing embeddings, but rather taking products of values of the embeddings.) Definition Let . A set of distinct embeddings of into a field is algebraically independent over if the only polynomial over for which is the zero function is the zero polynomial. Theorem 8.2.3 Let degree . Then be an infinite field, let hom is algebraically independent over , and therefore so is any nonempty subset of . hom Proof. Suppose that is a polynomial over for which for all . Let be a basis for over . Then and so be finite and separable of where and . However, Theorem 8.2.2 implies that is invertible and so any vector in has the form , for some , which shows that is zero on the infinite subfield of . Theorem 1.3.5 then implies that is the zero polynomial. *8.3 The Normal Basis Theorem
Let be a finite Galois extension of degree . Since is finite and separable, there exists a such that . As we know, the set A Field Extension as a Vector Space 207 is a basis for over . This type of basis is called a polynomial basis. A normal basis for over is a basis for over consisting of the roots of an irreducible polynomial over . We wish to show that any finite Galois extension has a normal basis. Theorem 8.2.2 can be reworded for finite Galois extensions as follows. Theorem 8.3.1 If is finite and Galois, with is a basis for over if and only if det Now, if is finite and Galois, it is simple and so the roots of min are then . . Moreover, Theorem 8.3.1 implies that this set is a (normal) basis for over if and only if det . To find such an element , consider the matrix For each , the product runs through and so each row of is a distinct permutation of to the columns of . Thus, we may write as runs through , . The same applies where for each , the row indices form a distinct permutation of and for each , the column indices form a distinct permutation of . Let be independent variables and consider the matrix We claim that the polynomial det is nonzero. Each row of is a distinct permutation of the variables and similarly for each column. Thus , is a permutation matrix, that is, each row and each column of , contains one and the rest 's. Since 208 Field Theory permutation matrices are nonsingular, we have , Hence, If . det , is an infinite field, Theorem 8.2.3 implies that the distinct embeddings of into are algebraically independent over and so there exists a for which det det Thus, we have proven the following. Theorem 8.3.2 If is an infinite field, then any finite Galois extension has a normal basis. This result holds for finite fields as well. The proof will be given in Chapter 9. Exercises
1. Let be finite. Prove that for all , Tr and Tr , Tr 2. Let be finite. Prove that if Tr , then and 3. If are finite and if Tr Tr hom show that Tr . If prove that 4. Let be finite and let 5. 6. State and prove a similar statement for the trace. Find a normal basis for the splitting field of over . If is finite and Galois, with , prove without appeal to Theorem 8.2.2, but rather using the Dedekind independence theorem, that if is a basis for over then det . A Field Extension as a Vector Space 209 7. Let min be a finite separable extension, with have degree . Show that
2 . Let 8. Let for over be finite and separable with form (8.2.1) and let be a basis . The dual basis to is a basis with the property that Tr
, where , dual bases if if and otherwise. In matrix terms, and are over is called a polynomial basis if it has the form for some . Any simple algebraic extension has a polynomial basis. Let be finite and separable, with polynomial basis . Let min Prove that the dual basis for is A basis for 9. If is a vector space, let denote the algebraic dual space of all linear functionals on . Note that if dim is finite then dim dim . a) Prove the Riesz Representation Theorem for nonsingular metric vector spaces: Let be a finitedimensional nonsingular metric vector space over and let be a linear functional on . Then there exists a unique vector such that for all . Hint: Let be defined by , . Define a map by . Show that is an isomorphism. b) Let be finite and separable, with form (8.2.1). Prove that for any linear functional there exists a unique for which Tr for all . Chapter 9 Finite Fields I: Basic Properties In this chapter and the next, we study finite fields, which play an important role in the applications of field theory, especially to coding theory, cryptology and combinatorics. For a thorough treatment of finite fields, the reader should consult the book Introduction to Finite Fields and Their Applications, by Lidl and Niederreiter, Cambridge University Press, 1986. 9.1 Finite Fields Redux
If is a field, then will denote the multiplicative group of all nonzero elements of . Let us recall some facts about finite fields that have already been established. Theorem 9.1.1 Let be a finite field. 1) has prime characteristic . (Theorem 0.4.4) 2) is cyclic. (Corollary 1.3.4) 3) Any finite extension of is simple. (Theorem 2.4.3) 4) is perfect, and so every algebraic extension of is separable and the Frobenius map is an automorphism of , for all . (Theorem 3.4.3) Lemma 9.1.2 If is a finite field and then . Proof. If is a basis for over , then each element of has a unique representation of the form , where . Since there are possibilities for each coefficient , we deduce that . Since a finite field 9.1.2 gives has prime characteristic , we have and so Lemma Corollary 9.1.3 If is a finite field with char for some positive integer . , then has elements 212 Field Theory From now on, unless otherwise stated, will represent a power of . will represent a prime number, and 9.2 Finite Fields as Splitting Fields
Let be a finite field of size . Then has exponent , that is, is a root of the polynomial has order and so every element . It follows that every element of Since , this polynomial has no multiple roots and so is precisely the set of roots of in some splitting field. In fact, since is a field, it is a splitting field for over the prime subfield . In symbols, Roots Split This has profound consequences for the behavior of finite fields. Existence
We have seen that every finite field of characteristic has elements for some . Conversely, let . If is the set of roots of , then is actually a field. For if , then and , whence and Thus . It follows that is a field and hence a splitting field for . Furthermore, since has no multiple roots, has size . Thus, for every prime power , there is a field of size . over , Of course, since each finite field of size is a splitting field for we know that all such fields are isomorphic. It is customary to denote a finite field of size by , or stands for Galois Field, in honor of Evariste Galois.) . (The symbol Theorem 9.2.1 1) Every finite field has size , for some prime and integer . 2) For every there is, up to isomorphism, a unique finite field of size , which is both the set of roots of and the splitting field for over . Finite Fields I: Basic Properties 213 Let us refer to the polynomial as the defining polynomial of the finite field . In view of this theorem, we will often refer to the finite field . An immediate consequence of the splitting field characterization of finite fields is that any extension of finite fields is normal. Corollary 9.2.2 The extension Hence, in the Galois correspondence for fields and all subgroups are closed. is a finite Galois extension. , all intermediate 9.3 The Subfields of a Finite Field
We wish to examine the subfields of a finite field . Note that if are positive integers and for , then and Hence, divides if and only if divides Repeating this shows that divides if and only if divides , that is, if and only if . In other words, . (9.3.1) over the prime subfield . ) The following are equivalent: divides the defining polynomial of Theorem 9.3.1 (Subfields of 1) 2) The defining polynomial of , that is, over the prime subfield . 3) Put another way, the following lattices are isomorphic (under the obvious maps): a) divides , under division b) divides , under division c) Subfields of , under set inclusion. Moreover, has exactly one subfield of size , for each . Proof. Two applications of (9.3.1) show that and so 1) and 2) are equivalent. Moreover, Roots Roots 214 Field Theory and so 2) and 3) are equivalent. For the last statement, if has two distinct subfields of size , then the polynomial would have more than roots in . 9.4 The Multiplicative Structure of a Finite Field
Since is cyclic, Theorem 0.2.11 implies the following theorem. of order . for each Theorem 9.4.1 There are exactly elements of and this accounts for all of the elements of It is customary to refer to any element of that generates the cyclic group as a primitive element of . However, this brings us into conflict with the term primitive as used earlier to denote any element of a field that generates the field using both field operations (addition and multiplication). Accordingly, we adopt the following definition. Definition Any element of that generates the cyclic group is called a group primitive element of . In contrast, if , then any element for which is called a field primitive element of over . Roots in a Finite Field
If , we may wish to know when when the equation has a th root in , that is, (9.4.1) has a solution in . This question has a simple answer in view of the fact that is cyclic. If is a group primitive element of then for some and so (9.4.1) has a solution if and only if for some integer , that is, that is, , which holds if and only if , for some integer . But this holds if and only if gcd Thus, equation (9.4.1) has a solution for all gcd , that is, if and only if and if and only if are relatively prime. Finite Fields I: Basic Properties 215 Theorem 9.4.2 1) Let be a group primitive element of if and only if gcd . Then has a th root in 2) Every element of has a th root if and only if and relatively prime, in which case every element has a unique th root. 3) The function are is a permutation of In this case, if and only if and are relatively prime. is called a permutation polynomial. 9.5 The Galois Group of a Finite Field
Since the extension over then is Galois, if is the Galois group of The structure of could not be simpler, as we now show. Theorem 9.5.1 The Galois group of over is cyclic of order , generated by the Frobenius automorphism . Proof. We have seen that the Frobenius map is an automorphism of . If , then and so fixes and is therefore in the Galois group . Moreover, the automorphisms are distinct elements of , for if then so , which implies that that . for all . Finally, since and , we see 9.6 Irreducible Polynomials over Finite Fields
Some of the most remarkable properties of finite fields stem from the fact that every finite field is not only the splitting field for the polynomial , but is also the set of roots of . This applies to the properties of irreducible polynomials over a finite field. Existence of Irreducible Polynomials
As to existence, if is a finite field and is a positive integer, then there is an irreducible polynomial of degree over . This follows from the fact that the extension is simple and so for min some . Then the minimal polynomial is irreducible of degree . 216 Field Theory The Splitting Field and Roots of an Irreducible Polynomial
Let be irreducible over of degree . Let be a root of is normal, it follows that splits in is a splitting field for . Thus, Moreover, . Since and so . and so the degree can be characterized as the smallest positive integer for which . Since the Galois group is the cyclic group , the roots of are Note that . can also be characterized as the smallest positive integer for which The Order of an Irreducible Polynomial
Since none of the roots of is zero, the roots belong to the multiplicative group . Moreover, since each root is obtained by applying an automorphism to a single root , all roots of have the same multiplicative order. Let us denote this order by . Thus, if and only if . The common order of the roots is referred to as the order of the irreducible polynomial and is denoted by . Note that this definition makes sense only for irreducible polynomials. As an aside, if the order of is , then each root of is group primitive, and we say that is primitive. Primitive polynomials play an important role in finite field arithmetic, as we will see in the next chapter. The Relationship Between Degree and Order
The relationship between the degree follows. First, note that and the order of can be gleaned as mod and since is the smallest positive integer for which the former holds, it is also the smallest positive integer for which the latter holds, that is, the order of modulo . It happens that this relationship between order and degree actually characterizes irreducibility. That is, if is a polynomial with root of order and if deg is equal to the order of modulo , then must be irreducible (in which case all roots have order ). For if is reducible, then is a root Finite Fields I: Basic Properties 217 of an irreducible factor of modulo . , with degree . Hence, is the order of Summary
Let us summarize. Theorem 9.6.1 For every finite field , and every positive integer , there exists an irreducible polynomial of degree over . Let be irreducible of order and let be a root of in some extension field. Let denote the order of in . 1) (Splitting Field) The splitting field of is . 2) if and only if . 3) (Roots) The roots of in a splitting field are and so is the smallest positive integer for which . 4) (Order of Roots) All roots of have order the same order , called the order of . 5) (Degree) The degree of is the smallest positive integer for which , or equivalently, . 6) (Relationship between degree and order characterizes irreducibility) Let be a polynomial over with order and degree . Then is irreducible if and only if Computing the Order of a Polynomial
To compute the order use the fact that of an irreducible polynomial of degree , we can and Let where the 's are distinct primes. Then where and, for each , 218 Field Theory if and only if , that is, if and only if Thus, is the smallest nonnegative integer for which . Example 9.6.1 Consider the irreducible polynomial . Since , we have over Let . Then is the smallest nonnegative integer for which Division shows that and so . For , division gives and so . . Thus, Thus , showing that is primitive over As another example, the polynomial irreducible over . If , then is also and so . Also,
7 7 and so . Thus, . Note that both of these polynomials have degree but they have different orders. This shows that the degree of an irreducible polynomial does not determine its order. *9.7 Normal Bases
Since any extension is simple, there is an which . Moreover, the set over . This type of basis is called a polynomial basis. for is a basis for Since the roots of an irreducible polynomial of degree over are distinct, it is natural to wonder whether there is an irreducible polynomial whose roots form a basis for over . Such a basis is referred to as Finite Fields I: Basic Properties 219 a normal basis. In short, a normal basis is a basis of roots of an irreducible polynomial. We saw in Chapter 8 that if is a finite Galois extension and is an infinite field, then has a normal basis over . This is also true for finite fields and stems from the fact that the members of the Galois group are linearly independent. Let be irreducible of degree over field of and the Galois group of is . Then is the splitting where is the Frobenius automorphism. But since these automorphisms are distinct, the Dedekind independence theorem tells us that they are linearly independent. This implies that as a linear operator on , the automorphism has minimal polynomial , for no polynomial of smaller degree can be satisfied by . But the characteristic polynomial of is monic, has degree and is divisible by the minimal polynomial (this is the CayleyHamilton theorem), and so it is also equal to . The following result from linear algebra, which we will not prove here, is just what we need. Theorem 9.7.1 Let vector space over a field be a linear operator on a finitedimensional . Then contains a vector for which , is a basis for if and only if the minimal polynomial and characteristic polynomial of are equal. This theorem implies that there is an Roots is a (normal) basis for over . for over for which Theorem 9.7.2 There exists a normal basis . *9.8 The Algebraic Closure of a Finite Field
In this section, we determine the algebraic closure of a finite field . Since is algebraic for all positive integers , an algebraic closure of must contain all of the fields . 220 Field Theory Since , it follows that and so the union is an extension field of that contains , for all . Moreover, if is a field for which for all , then , that is, is the smallest field containing each . Theorem 9.8.1 The field is the algebraic closure of . Proof. Every element of lies in some , whence it is algebraic over . Thus is algebraic over . Now let be an irreducible polynomial over of degree . Then the coefficients of lie in some and so is irreducible as a polynomial over . Hence, the splitting field for is and so splits over . Steinitz Numbers
We wish now to describe the subfields of the algebraic closure . Recall that a field is a subfield of if and only if where . The set of positive integers is a complete lattice where and gcd lcm . If we denote by the set of all finite fields or more properly the set of all isomorphism classes of finite fields) that contain , then is also a complete lattice where and . Theorem 9.8.2 The map defined by is an orderpreserving bijection. Hence, it is an isomorphism of lattices, that is, 1) if and only if 2) 3) Proof. Left to the reader. It is clear that the lattice of intermediate fields between and is isomorphic to the sublattice of consisting of all positive integers dividing . In order to describe the lattice of intermediate fields between and , we make the following definition. Definition A Steinitz number is an expression of the form Finite Fields I: Basic Properties 221 where is the th prime and . We denote the set of all Steinitz numbers by . Two Steinitz numbers are equal if and only if the exponents of corresponding prime numbers are equal. We will denote arbitrary Steinitz numbers using uppercase letters and reserve lowercase letters strictly for ordinary positive integers. We will take certain obvious liberties when writing Steinitz numbers, such as omitting factors with exponent equal to . Thus, any positive integer is a Steinitz number. We next define the arithmetic of Steinitz numbers. Definition Let and 1) The product and quotient of be Steinitz numbers. are defined by and and where 2) We say that It is clear that . Also, . divides and write if for all . for all positive natural numbers if and only if if and only if and . Theorem 9.8.3 Under the relation of "divides" given in the previous definition, the set is a complete distributive lattice, with meet and join given by
min and max Moreover, the set of positive integers is a sublattice of . Subfields of the Algebraic Closure
We can now describe the subfields of subfields of that contain . Definition If is a Steinitz number, let . Let denote the lattice of all where, as indicated by the lowercase notation, If . Thus subfield of then where containing . for some lcm is a positive integer. and . It follows that for some is a 222 Field Theory Theorem 9.8.4 The map defined by is an orderpreserving bijection. Hence, it is an isomorphism of lattices, that is, 1) if and only if , 2) , 3) . In addition, is finite if and only if is a positive integer. Proof. We begin by showing that if and only if . One direction follows immediately from the definition: if then . Suppose that . Let be a field primitive element of over . Then and so for some . Hence , which implies that , whence . Since if and only if , it follows that if and only if that is, if and only if To see that is injective, if the other and so if and only if To see that which which is surjective, let . For each prime . , then each field is contained in , which implies that . . We must find an be the largest power of for for (9.8.1) , let where if (9.8.1) holds for all positive integers . Let We claim that (9.8.2) The second equality is by definition and the first field is clearly contained in the second. Also, if , then where and so Finite Fields I: Basic Properties 223 It follows that (9.8.2) holds. This implies that For the reverse inclusion, if then for some . If then and so and so for all , by the maximality of . Hence . This shows that . Hence is surjective. We leave the rest of the proof to the reader. and so Since the largest Steinitz number is this corresponds to the largest subfield of , that is, Exercises
1. 2. Determine the number of subfields of and . Group primitive elements of , prime, can often be found by experimentation and the fact that if and and then . For instance, if , then by checking some small primes, we see that and , whence and so is group primitive for . a) For , show that and . Find an element of order to pair with . b) If is group primitive for , an odd prime, then what is ? c) Prove Wilson's theorem If is an odd prime then mod Hint: The left side is the product of all nonzero elements in . Conisder this product from the point of view of a group primitive element . Show that except for the case of , the sum of all the elements in a finite field is equal to . Find all group primitive elements of . 3. 4. 224 Field Theory 5. 6. Show that the polynomial Is it primitive? Let be an arbitrary field. Prove that if field. is irreducible over is cyclic then . must be a finite Find the order of the following irreducible polynomials. 7. over . 8. over . 9. over . 10. over . 11. over . 12. over . 13. over . 14. over . 15. Show that every element in has a unique th root, for . 16. If , show that exactly onehalf of the nonzero elements of have square roots. 17. Show that if and is a positive integer, then divides . 18. Find a normal basis for over . Hint: Let be a root of the irreducible polynomial . 19. Show that . 20. Let be any strictly increasing infinite sequence of positive integers. Prove that . 21. Show that . 22. Let be a field satisfying . Show that all the proper subfields of are finite if and only if is finite or where for some prime . 23. Show that has no maximal subfields. 24. Show that is not finite for any proper subfield . 25. Show that has an uncountable number of nonisomorphic subfields. 26. Let . Show that is finite if and only if is finite, in which case the two numbers are equal. Chapter 10 Finite Fields II: Additional Properties 10.1 Finite Field Arithmetic
There are various ways in which to represent the elements of a finite field. Since every finite field is simple, it has the form for some and so the elements of are polynomials in of degree less than deg . Another way to represent the elements of a finite field is to use the fact that is cyclic, and so its elements are all powers of a group primitive element. It is clear that addition is more easily performed when field elements are written as polynomials and multiplication is more easily performed when all elements are written as a power of a single group primitive element. Fortunately, the two methods can be combined to provide an effective means for doing finite field arithmetic. Example 10.1.1 Consider the finite field The polynomial as an extension of . is irreducible over . To see this, note that if is reducible, it must have either a linear or a quadratic factor. But since and , it has no linear factors. To see that has no quadratic factors, note that there are precisely four quadratic polynomials over , namely, and it is easy to check that no product of any two of these polynomials equals . Thus, letting be a root of , we can represent the elements of the binary polynomials of degree or less in , as follows: as 226 Field Theory Constant: Linear: Quadratic: Cubic: , , , , , Addition of elements of is quite simple, since it is just addition of polynomials, but multiplication requires reduction modulo , using the relation . On the other hand, observe that
5 + + and so primitive and . Since and , we conclude that
4 is group With this representation, multiplication is all but trivial, but addition is cumbersome. We can link the two representations of by computing a table showing how each element can be represented as a polynomial in of degree at most . Using the fact that , we have and so on. The complete list, given in Table 10.1.1 , is known as a field table for . As is customary, we write only the exponent for , and for the polynomial . Finite Fields II: Additional Properties 227 Table 10.1.1 Computations using this table are quite straightforward; for example, 0 9 9 Thus, the key to doing arithmetic in a finite field is having a group primitive element, along with its minimal (primitive) polynomial. In general, the task of finding primitive polynomials is not easy. There are various methods that achieve some measure of success in certain cases, and we mention one such method at the end of Section 11.2. Fortunately, extensive tables of primitive polynomials and field tables have been constructed. Let us use the primitive polynomial and the field table for to compute the minimal polynomial over for each element of . We begin by computing sets of conjugates, using Theorem 9.6.1 and the fact that 6 , Conjugates of : Conjugates of : Conjugates of : Conjugates of : Letting be the minimal polynomial for , we have, for example + 228 Field Theory The field table for gives and since , we have The other minimal polynomials are computed similarly. The complete list is 6 Being able to factor polynomials of the form is important for a variety of applications of finite field theory, especially to coding theory. Since the roots of over are precisely the elements of , we have Of course, in order to obtain this factorization, we worked in the splitting field . Let us turn to a method for factoring polynomials over that does not require working in any extension of . Factoring over : Berlekamp's Algorithm Berlekamp's algorithm is an algorithm for factoring polynomials over . Suppose that is a polynomial over of degree . Let us first show that we can reduce the problem of factoring to one of factoring a polynomial with no repeated factors. We know that has a repeated factor if and only if common factor. Write gcd Let gcd then . If and so gcd then has no repeated factors. If and have a and we can factor (or repeat the process). Otherwise, is a nonconstant polynomial with degree less than that of and has no repeated factors. Thus, we can consider the polynomials and separately. For the former polynomial, we can repeat the above argument until the factoring problem reduces to one of factoring polynomials with no repeated factors. Finite Fields II: Additional Properties 229 So let us suppose that is the product of distinct irreducible factors. (Actually, the factoring algorithm that we are about to describe does not require this restriction on , but the formula for the number of irreducible factors that we will present does.) Suppose that we can find a nonconstant polynomial for which of degree less than Since is the set of roots of , we have and so Also, if in general, then , where gcd gcd are pairwise relatively prime, Hence, since the polynomials , we have gcd gcd are pairwise relatively prime for gcd Note that the degree of each of these factors is at most deg and so this factorization of is nontrivial. Note also that the Euclidean algorithm can be used to find the gcd of the pairs of polynomials in the previous factorization and so if we can find such a polynomial , then we will have an algorithm for finding a nontrivial factorization of . A polynomial for which is called an reducing polynomial. We are interested in nonconstant reducing polynomials with degree less than the degree of , since these polynomials provide factorizations of . To find such an reducing polynomial , write Then since we are working over a field of characteristic , and since modulo , it follows that and so 230 Field Theory Now suppose that is divisible by , where deg if and only if . Then but since the right hand sum has degree less that that of to , this is equivalent and this is equivalent to a system of linear equations. To express this system in matrix form, suppose that . Then the previous equation is equivalent to the system for . In matrix terms, if , and is the row matrix of coefficients of , then this system is Example 10.1.1 Consider the polynomial over . First, we find the polynomials by dividing by , to get Finite Fields II: Additional Properties 231 Hence, and our system is whose solution is arbitrary; The only nonconstant solution is ; where . It follows that and so, using Euclid's algorithm for the gcd, we get the factorization gcd gcd gcd gcd The Number of Irreducible Factors Knowledge of the number of irreducible factors of would help us determine when the factorization algorithm has produced a complete factorization of into irreducible factors. Suppose that where the are distinct monic, irreducible polynomials over . Let be the set of reducing polynomials with degree less than that of , including the constant polynomials. Note that is isomorphic to the null space null of the matrix of the Berlekamp algorithm. 232 Field Theory If , then and since the polynomials on the right are relatively prime, each divides precisely one of these polynomials, say . This is a system of congruences mod mod and since the 's are relatively prime (this is where we use the fact that the are distinct), the Chinese remainder theorem tells us that there is a unique solution modulo , that is, a unique solution of degree less than that of . In other words, there is at most one reducing polynomial for each tuple . But if is a solution to this system, then for all and so , whence . It follows that there is precisely one reducing polynomial for each tuple in . Hence,
dim null rk that is, the number of distinct irreducible factors of rk is Example 10.1.2 The matrix from Example 10.1.1 has rank , which can be determined by applying elementary row operations to reduce the matrix to echelon form. Hence, the nullity is and so the factorization in that example is complete. *10.2 The Number of Irreducible Polynomials
Of course, if is a finite field, then there is only a finite number of polynomials of a given degree over . It is possible to obtain an explicit formula for the number of irreducible polynomials of degree over by using Mo bius inversion. (See the appendix for a discussion of Mo bius inversion.) First, we need the following result. Theorem 10.2.1 Let be a finite field, and let be a positive integer. Then the product of all monic irreducible polynomials over , whose degree divides is Proof. According to Theorem 9.6.1, an irreducible polynomial divides Finite Fields II: Additional Properties 233 if and only if deg . Hence, is a product of irreducible polynomials whose degrees divide and every irreducible polynomial whose degree divides divides . Since no two such irreducible polynomials have any roots in common and since has no multiple roots, the result follows. Let us denote the number of monic irreducible polynomials of degree over by . By counting degrees, Theorem 10.2.1 gives the following. Corollary 10.2.2 For all positive integers and , we have Now we can apply Mo bius inversion to get an explicit formula for Classical Mo bius inversion is . (10.2.1) where the Mo bius function is defined by if if otherwise for distinct primes Corollary 10.2.3 The number degree over is of monic irreducible polynomials of Proof. Letting and in (10.2.1) gives the result. Example 10.2.1 The number of monic irreducible polynomials of degree over is
2 2 The number of monic irreducible polynomials of degree over is as we would expect from the results of Example 10.1.1. 234 Field Theory Mo bius inversion can also be used to find the product of all monic irreducible polynomials of degree over . Let us denote this product by . Then Theorem 10.2.1 is equivalent to Applying the multiplicative version of Mo bius inversion gives the following. Corollary 10.2.4 The product degree over is of all monic irreducible polynomials of Example 10.2.2 For
6 6 and , we get
4 5 2 9 *10.3 Polynomial Functions
Finite fields have the special property that any function from a finite field to itself can be represented by a polynomial. As a matter of fact, this property actually characterizes finite fields from among all commutative rings (finite and infinite) Since has size , there are precisely functions from to itself. Among these functions are the polynomial functions where . We will denote this polynomial function by as well. If and are polynomial functions on then as functions if and only if for all , which holds if and only if Thus, two polynomials represent the same function if and only if they are congruent modulo . Since every polynomial is congruent modulo to precisely one polynomial of degree less than (namely, its remainder after dividing by , and since there are polynomials of degree less than , we have the following theorem. (Proof of the last statement in part 2 of the theorem is left to the reader.) Theorem 10.3.1 1) Two polynomials over represent the same polynomial function on if and only if they are congruent modulo . Finite Fields II: Additional Properties 235 2) Every function is a polynomial function, for a unique polynomial of degree less than . In fact, the unique polynomial of degree less than that represents is (The representation of given in part 2) above is the Lagrange interpolation formula as applied to finite fields.) Part 2) has a very interesting converse as well. Theorem 10.3.2 If is a commutative ring and if every function is a polynomial function, that is, for some , then is a finite field. Proof. First, we show that is finite. Suppose that . The number of functions from to itself is and the number of polynomials over is the same as the number of finite sequences with elements from , which is . Since distinct functions are represented by distinct polynomials, we must have , which happens only when is finite. Thus, is a finite set. Now let with . Define a function if if By hypothesis, there exists a polynomial for which by and , for Setting gives and so Thus, we conclude that for any and any , there is a for which . In other words, the map defined by is surjective. Since is a finite set, must also be injective. Hence, , implies that and so has no zero divisors. In addition, since is surjective, there exists a for which , that is, . If then and since is commutative and has no zero divisors, we may cancel to get . Thus is the multiplicative identity of . Hence is a finite integral domain, that is, a finite field. 236 Field Theory *10.4 Linearized Polynomials
We now turn to a discussion of linear operators on over . We will see that all such linear operators can be expressed as polynomial functions of a very special type. Definition A polynomial of the form with coefficients polynomial, over is called a linearized polynomial, or a .  The term linearized polynomial comes from the following theorem, whose proof is left to the reader. Theorem 10.4.1 Let and be a linearized polynomial over , then . If Thus, the polynomial function over . is a linear operator on The roots of a polynomial in a splitting field have some rather special properties, which we give in the next two theorems. Theorem 10.4.2 Let be a nonzero polynomial over , with splitting field . Then each root of in has the same multiplicity, which must be either or else a power of . Furthermore, the roots of form a vector subspace of over . Proof. Since , if then all roots of are simple. On the other hand, suppose that but . Then since , we have and so which is the th power of a linearized polynomial with nonzero constant term, and therefore has only simple roots. Hence, each root of has multiplicity . We leave proof of the fact that the roots form a vector subspace of to the reader. Finite Fields II: Additional Properties 237 The following theorem, whose proof we omit, is a sort of converse to Theorem 10.4.1. (For a proof of this theorem, and more on polynomials, see the book by Lidl and Niederreiter (1986).) Theorem 10.4.3 Let be a vector subspace of any nonnegative integer , the polynomial over . Then for is a polynomial over If . is a polynomial, then as a function, we have where is the Frobenius automorphism. Thus, as an operator Since we may reduce the expression for to a polynomial in of degree at most . In fact, adding coefficients if necessary, we can say that every polynomial function on has the standard form for . There are such polynomial functions on , and this happens also to be the number of linear operators on over . Moreover, since the maps are linearly independent over , we deduce that each polynomial in standard form represents a unique linear operator. Thus, we have characterized the linear operators on over . Theorem 10.4.4 Every linear operator on over represented by a unique polynomial in standard form can be for some . 238 Field Theory Exercises
1. 2. Construct two distinct finite field tables for Factor the polynomial over . 3. over Factor 4. 5. 6. over Factor Calculate Show that over . . and
2 2 7. 2 Hence, . Finally, show that Show that the unique polynomial of degree less than function is . that represents the Prove that a linearized polynomial over is a linear operator on over . 9. Prove that the roots of a polynomial over form a vector subspace of the splitting field over . 10. Prove that the greatest common divisor of two polynomials over is a polynomial, but the least common multiple need not be a polynomial. 8. Chapter 11 The Roots of Unity Polynomials of the form , where , are known as binomials. Even though binomials have a simple form, their study is quite involved, as is evidenced by the fact that the Galois group of a binomial is often nonabelian. As we will see, an understanding of the binomial is key to an understanding of all binomials. We will have use for the following definition. Definition The exponent characteristic expchar be if char and char otherwise. of a field is defined to 11.1 Roots of Unity
The roots of the binomial over a field are referred to as the th roots of unity over . Throughout this section, we will let be a field with expchar , a splitting field for over and the set of th roots of unity over , located in . Note that if then and so the th roots of unity are the same as the th roots of unity, taken with a higher multiplicity. Thus, from now on, we assume that . Theorem 11.1.1 The set of th roots of unity over is a cyclic group of order under multiplication. Moreover, if then
mn where the product of groups is direct. Proof. Clearly implies . Hence, is a finite subgroup of the multiplicative group of nonzero elements of the field . By Corollary 1.3.4, is cyclic. Since , we have 240 Field Theory showing that is separable, and so then . and since , whence For the second part, if there exist such that which shows that are distinct and since . Hence, the products in the group , it follows that . Definition An element of order , that is, a generator of , is called a primitive th root of unity over . We shall denote the set of all primitive th roots of unity over by and reserve the notation for a primitive th root of unity. Note that a primitive th root of unity , being a group primitive element, is also a field primitive element of , that is However, in general, roots of unity. Theorem 11.1.2 1) If then has field primitive elements that are not primitive th and . Hence, there is a bijection from onto the abelian (but not necessarily cyclic) group of all elements of that are relatively prime to , that is, to the group of units of . 2) If then . 3) If then . Proof. Part 1) follows from the fact that if then if and only if . For part 2), if then and so order , the set . Thus . For the reverse inclusion, since has consists of element of distinct roots of unity of order belongs to , since and so . But each Roots of Unity 241 and so For part 3), since . lcm we have and so are distinct, so are the products in . Now, since the products in . Hence which shows that . 11.2 Cyclotomic Extensions
The term cyclotomy is the process of dividing a circle into equal parts, which is precisely the effect obtained by plotting the th roots of unity over in the complex plane. Definition Let be a field. A splitting field cyclotomic extension of order of . Since of over is called a for is the splitting field of a separable polynomial, it follows that is a finite Galois extension and deg min Now, any is uniquely determined by its value on a fixed , and since preserves order, must be one of the primitive roots of unity in , that is, where . Since it follows that mod and so the map that , the map subgroup of . is a homomorphism. Since implies is a monomorphism and thus is isomorphic to a 242 Field Theory Theorem 11.2.1 If is a cyclotomic extension of order , then isomorphic to a subgroup of , the group of units of . Hence, abelian and . is is Since the structure of is clearly important, we record the following theorem, whose proof is left as an exercise. Theorem 11.2.2 Let , where the 's are distinct primes. Then Moreover, prime. is cyclic if and only if or , where is an odd Corollary 11.2.3 A cyclotomic extension is abelian and if , where is an odd prime, then is cyclic. , or Cyclotomic Polynomials
To investigate the properties of cyclotomic extensions further, we factor the polynomial . Since each root of this polynomial is a primitive th root of unity for some , we define the th cyclotomic polynomial to be the polynomial whose roots are precisely the primitive th roots of unity. Thus, if is a primitive th root of unity, then It follows that deg and since each side is the product of the linear factors , as varies over all th roots of unity. Note that cyclotomic polynomials are not necesssarily irreducible, and we will explore this issue as soon as we have recorded the basic properties of these polynomials. Note also that the cyclotomic polynomial where expchar . Theorem 11.2.4 Let 1) deg is defined only for be the th cyclotomic polynomial over . . Roots of Unity 243 2) The following product formula holds: (11.2.1) 3) is monic and has coefficients in the prime subfield of 4) If then the coefficients of are integers. 5) The cyclotomic polynomials are given by . where is the Mobius function, defined by if if otherwise for distinct primes Note that some of the exponents may be equal to , and so a little additional algebraic manipulation may be required to obtain as a product of polynomials. Proof. Parts 3) and 4) can be proved by induction, using formula (11.2.1). In particular, let be the prime subfield of . It is clear from the definition that is monic. Since , the result is true for . If is a prime then
2 and the result holds for of . Then . Assume that 3) and 4) hold for all proper divisors By the induction hypothesis, has coefficients in , and therefore so does . Moreover, if , then has integer coefficients and since is monic (and therefore primitive), Theorem 1.2.2 implies that has integer coefficients. Part 5) follows by Mo bius inversion. (See the appendix for a discussion of Mo bius inversion.) Example 11.2.1 Formula (11.2.1) can be used to compute cyclotomic polynomials rather readily, starting from the fact that and
2 244 Field Theory for prime. Thus, for example, 6 and
5 5 5 This gives us, for instance, the following factorization of cyclotomic polynomials: into The Mo bius inversion formula gives Part 4) of Theorem 11.2.4 describes a factorization of within the prime subfield of . In general, however, this is not a prime factorization since is not irreducible. For instance, comparing Examples 11.2.1 and 10.1.1 shows that 5 is reducible over . When Is the Galois Group as Large as Possible?
We have seen that if is a cyclotomic extension of order , then isomorphic to a subgroup of , which has order . Thus, isomorphic to the full group if and only if is is that is, if and only if the cyclotomic polynomial case min . is irreducible, in which Theorem 11.2.5 Let be the splitting field for over . Then isomorphic to if and only if the th cyclotomic polynomial irreducible over , in which case . min is is Roots of Unity 245 The Irreducibility of Cyclotomic Polynomials
With regard to the irreducibility of cyclotomic polynomials, we have the following important results. In particular, if is irreducible, then so is for . Also, over the rational numbers, all cyclotomic polynomials are irreducible. Note that since the Galois group of is isomorphic to a subgroup of , which has order , it follows that the degree of divides . Theorem 11.2.6 Let be irreducible over and let , where . Then is also irreducible over . As usual, we assume that char . Proof. Let be a prime and consider the tower The first step has degree . But extension is and so and the second step has degree , since irreducible implies that the degree of the full with If and then . and we have with and so divide that and . It follows that is irreducible. If then for . If , then and so . If , where then , which does not is odd. It follows and so Finally, if is irreducible. then and so with and . Hence, and again is irreducible. Thus, we have shown that if with prime, then is irreducible. Suppose that , that is, . Then repeatedly applying the argument above shows that is irreducible. 246 Field Theory Theorem 11.2.7 All cyclotomic polynomials over the rational field are irreducible over . Therefore, and . Proof. Suppose that is a nontrivial factorization, where we may assume that both factors are monic and have integer coefficients. Assume that is irreducible and that is a root of . We show that is also a root of , for any prime . For if not, then must be a root of . Hence, is a root of , which implies that and where is monic and has integer coefficients. Since mod , for any integer , we have mod and so taking residues gives mod or, in a different notation in . It follows that and have a common root in some extension of . However, , which has no multiple roots in any extension. This contradiction implies that is a root of . is a root of , then so is , where . If is a prime and , the same argument applied to shows that is also a root of . In fact, for any , it follows that is a root of , that is, all roots of are roots of , and so , whence is irreducible over . Thus, if Finite Fields
If the base field extension cyclic with generator is a finite field, then we know that the cyclotomic is also a finite field and the Galois group is : Since the order of is , Theorem 9.6.1 implies that From this, we also get a simple criterion to determine when a cyclotomic polynomial is irreducible. Theorem 11.2.8 Let be the splitting field for . Then 1) 2) 3) is isomorphic to the cyclic subgroup over , where of . Roots of Unity 247 4) The following are equivalent a) b) c) The cyclotomic polynomial Let us consider an example. Example 11.2.2 Since is irreducible over . the polynomial is irreducible over and has degree . Since the polynomial of degree is not irreducible over . Types of Primitivity There are three types of elements in the splitting field of over a finite field that are referred to as primitive: field primitive elements, group primitive elements and primitive roots of unity. Since each type of primitive element is field primitive, that is, , each type of primitive element has degree . However, the orders of each type of primitive element differ. If is field primitive, that is, 1) deg 2) If is group primitive, that is, 1) deg 2) , then , then If is a primitive th root of unity, that is, 1) deg 2) Given a group primitive element , then of , we can identify from among its powers which are the primitive th roots of unity. In fact, unity if and only if is a primitive th root of In general, the equation 248 Field Theory is equivalent to , or which holds if and only if and we have the following. where . In this case, Theorem 11.2.9 Let be a group primitive element of the cyclotomic extension of order . Then is a primitive th root of unity if and only if where and . More on Cyclotomic Polynomials If is monic and irreducible over and has order , then each root of has order and thus . Since every monic irreducible factor of has order , and since these factors have no common roots, we conclude that is the product of all monic irreducible polynomials of order . According to Theorem 9.6.1, the degree of any such factor is . Hence, the number of monic irreducible polynomials of order is . Theorem 11.2.10 Let be a positive integer. 1) The cyclotomic polynomial over is the product of all monic irreducible polynomials of order over . 2) The number of monic irreducible polynomials over of order is , where is the order of mod . Let us mention that the roots of the st cyclotomic polynomial have order and so are group primitive elements of . In other words, the monic irreducible factors of are precisely the primitive polynomials of over . Thus, one way to find primitive polynomials is to factor this cyclotomic polynomial. Example 11.2.3 We have at our disposal a number of tools for factoring polynomials of the form over , for : 1) 2) Roots of Unity 249 3) For prime,
2 4) 5) is irreducible over if and only if . 6) Over , the polynomial is the product of all monic irreducible polynomials of order over . 7) A polynomial over is irreducible if and only if its order is . 8) Let . Then is irreducible if and only if is irreducible. Moreover, if and are roots of an irreducible polynomial , then is also irreducible and Hence, translation by an element of the base field preserves the property of being conjugate (that is, being roots of the same irreducible polynomial). To illustrate, consider the polynomial . Over , we have A small table of order/degrees is useful: This table shows that and are irreducible, but that is not. However, since the roots of have order , the degree of any irreducible factor of must satisfy . Thus, factors into a product of two irreducible quartics, which are primitive polynomials for . To find the quartic factors of quartic factors must have the form , we can proceed by brute force. The 250 Field Theory where implies that either . Since , we must have or exactly one of or is . , which If , then , which is not a factor of , because the orders are not equal. Hence, exactly one of or is equal to . If , then which is not irreducible. Hence, we are left with only two possibilities, and where the factors are irreducible over Another approach is to observe that . is irreducible and so therefore is and since and so does not divide . or , its roots have order Once we have factored , we can find a group primitive element of its splitting field, which is . In particular, a group primitive element has order , and so is a root of . So let be a root of the irreducible polynomial over . Then where primitive . Note that is also a primitive th roots being , where th root of unity, the other . *11.3 Normal Bases and Roots of Unity
Recall that a normal basis for is a basis for over that consists of the roots of an irreducible polynomial over . We have seen that in some important cases (especially ), the cyclotomic polynomials are irreducible over , which leaves open the possibility that the primitive th roots of unity might form a normal basis for over . Indeed, if is irreducible then and so min deg and since the roots of are distinct, there is the right number of primitive th roots of unity and they will form a basis for over if and only if they span over . Roots of Unity 251 Theorem 11.3.1 Let be a field with the property that is irreducible over for all . Then is a normal basis for the cyclotomic extension if and only if is the product of distinct primes. Proof. First, let be prime and . Consider the extension Since is irreducible, it follows that so the powers is irreducible over and form a basis for over . But and so the set . is a normal basis for over Now we can proceed by induction on . We have just seen that the result is true for prime. Suppose that , where . Then by the inductive hypothesis, we may assume that is a normal basis for over . Then the product is a basis for over . But and so is a normal basis for over . For the converse, let for . Since (an exercise) the coefficient of in is , whence the sum of the roots of , that is, the sum of the primitive th roots of unity, is , showing that these roots are linearly dependent. Hence, they cannot form a basis for over . *11.4 Wedderburn's Theorem
In this section, we present an important result whose proof uses the properties of cyclotomic polynomials. Theorem 11.4.1 (Wedderburn's Theorem) If is a finite division ring, then is a field. Proof. Let the multiplicative group act on itself by conjugation. The stabilizer of is the centralizer and the class equation is 252 Field Theory where the sum is taken over one representative from each conjugacy class of size greater than . If we assume for the purposes of contradiction that , then the sum on the far right is not an empty sum and for some . The sets for all and are subrings of and, in fact, field. Let . Since vector spaces over and so is a commutative division ring; that is, a , we may view and as and for integers . The class equation now gives and since If But , it follows that . . , is the th cyclotomic polynomial over , then divides also divides each summand on the far right above, since for , we have and divides the righthand side. It follows that other hand, . On the and since and implies that is commutative, that is, , we have a contradiction. Hence is a field. Roots of Unity 253 *11.5 Realizing Groups as Galois Groups
A group is said to be realizable over a field if there is an extension whose Galois group is . Since any finite group of order is isomorphic to a subgroup of a symmetric group , we have the following. Theorem 11.5.1 Let be a field. Every finite group is realizable over some extension of . Proof. Let be a group of order . Let be algebraically independent over and let be the elementary symmetric polynomials in the 's. Then is a Galois extension whose Galois group is isomorphic to . (See Theorem 7.2.3.) We may assume that is a subgroup of and since is closed in the Galois correspondence, it is the Galois group of fix . It is a major unsolved problem to determine which finite groups are realizable over the rational numbers . We shall prove that any finite abelian group is realizable over . It is also true that for any , the symmetric group is realizable over , but we shall prove this only when is a prime. Realizing Finite Abelian Groups over
We wish to show that any finite abelian group is realizable over the rational field . Since all cyclotomic polynomials are irreducible over the rationals, the extension has Galois group , which is finite and abelian. For any subgroup we have the corresponding tower of fields fix and since the extension is Galois and all subgroups are normal, the quotient is the Galois group of the extension fix . Hence, it is sufficient to show that any finite abelian group is isomorphic to a , for some . Since is finite and abelian, we have quotient where is cyclic of degree . If we show that is isomorphic to a quotient of the form , where the 's are distinct odd primes, then if and , it follows that 254 Field Theory as desired. Now, if is an odd prime, then is cyclic of order and so all we need to do is find distinct odd primes for which , because a cyclic group of order has quotient groups of all orders dividing . Put another way, we seek a set of distinct primes of the form , for . It is a famous theorem of Dirichlet that there are infinitely many primes of the form provided that and so the case is what we require. First a lemma on cyclotomic polynomials. Lemma 11.5.2 Let be a prime and let . Let be the polynomial obtained from by taking the residue of each coefficient modulo . Then is the th cyclotomic polynomial over . Proof. Let be the th cyclotomic polynomial over . If is a prime then and are all equal to
2 and so the result holds for prime. Let all proper divisors of . Since and suppose the result holds for taking residues modulo gives over . But over . and since for all , , it follows that Theorem 11.5.3 Let be a positive integer. Then there are infinitely many prime numbers of the form , where is a positive integer. Proof. Suppose to the contrary that is a complete list of all primes of the form . Let . Let be the th cyclotomic Roots of Unity 255 polynomial over and consider the polynomial . Since has integer coefficients, is an integer for all . Since can equal , or for only a finite number of positive integers , there exists a positive integer for which . . Let be a prime dividing Since , we have which implies that , hence contradiction, we show that has the form for . , then . To arrive at a If is the th cyclotomic polynomial over previous lemma imply that and the in , where the overbar denotes residue modulo . Thus, root of unity over Z , that is, has order in . Hence, is a primitive th and so , that is, has the form . We can now put the pieces together. Theorem 11.5.4 Let be a finite abelian group. Then there exists an integer and a field such that , where and such that . Realizing over We begin by discussing a sometimes useful tool for showing that the Galois group of a polynomial is a symmetric group. Let be the Galois group of an irreducible polynomial over , thought of as a group of permutations on the set of roots of . Then acts transitively on . Let us define an equivalence relation on by saying that if and only if either or the transposition is an element of . It is easy to see that this is an equivalence relation on . Let be the equivalence class containing . Suppose that contains a transposition . Then for any , we have In other words, if then and so have the same cardinality and since equivalence classes have the same cardinality. . It follows that acts transitively on and , all 256 Field Theory Hence, if has a prime number of roots, then there can be only one equivalence class, which implies that is in for all . Since contains every transposition, it must be the symmetric group on . We have proved the following. Theorem 11.5.5 If is a separable polynomial of prime degree and if the Galois group of contains a transposition, then is isomorphic to the symmetric group . Corollary 11.5.6 If is irreducible of prime degree and if has precisely two nonreal roots, then the Galois group of is isomorphic to the symmetric group . Proof. Let be a splitting field for over . Complex conjugation is an automorphism of leaving fixed. Moreover, since is normal, . Since leaves the real roots of fixed, is a transposition on the roots of . Thus, the theorem applies. Example 11.5.1 Consider the polynomial , which is irreducible over by Eisenstein's criterion. A quick sketch of the graph reveals that has precisely real roots and so its Galois group is isomorphic to . Corollary 11.5.6 is just what we need to establish that is realizable over . Theorem 11.5.7 Let be a prime. There exists an irreducible polynomial over of degree such that has precisely two nonreal roots. Hence, the symmetric group is realizable over . Proof. The result is easy for and , so let us assume that . Let be a positive integer and be an odd integer. Let be distinct even 2 integers and let
2 It is easy to see from the graph that Moreover, if is an odd integer, then has relative maxima. Let and since real roots. . Since the relative maxima of are all greater than and , we deduce that has at least We wish to choose a value of for which has at least one nonreal root , for then the complex conjugate is also a root, implying that has two nonreal roots and real roots. Let the roots of in a splitting field be Roots of Unity 257 . Then Equating coefficients of and and gives and so If is sufficiently large, then must be nonreal, as desired. is negative, whence at least one of the roots It is left to show that criterion. Let us write is irreducible, which we do using Eisenstein's In the product , each coefficient except the leading one is divisible by . Hence, we may write Multiplying by gives Taking to be even, we deduce that all nonleading coefficients of are even. In addition, the constant term of is divisible by since . It follows that is monic, all nonleading coefficients are divisible by , but the constant term is not divisible by . Therefore is irreducible and the proof is complete. Exercises
All cyclotomic polynomials are assumed to be over fields for which they are defined. 1. Prove that if where then . 258 Field Theory 2. 3. 4. 5. 6. 7. 8. When is a group primitive element of the cyclotomic extension also a primitive th root of unity over ? If , how many th roots of unity are there over ? What is the splitting field for over ? Find the primitive th roots of unity in this splitting field. Do the same for the th roots of unity over . If are the th roots of unity over show that for . What about when ? If , prove that is irreducible over if and only if is prime and is irreducible. Show that if is a prime, then . Show that if . Verify the following properties of the cyclotomic polynomials. As usual, is a prime number. 9. for . 10. for all . 11. 12. If is the decomposition of into a product of powers of distinct primes, then 13. 14. 15. Evaluate for . . for . On the structure of
16. If where .
are distinct prime powers then 17. Let be prime and let . a) Show that . b) Show that has an element of order . Hint: consider an element of order modulo , which exists since is a field. Compute the order of as an element of . c) Show that has order . Hint: Show that if then where . Then consider the powers d) Show that is cyclic. e) Show that is cyclic if and only if f) Show that is cyclic if and only if , etc. or . , or . Roots of Unity 259 18. Prove that if then there exists an irreducible polynomial of degree over whose Galois group is isomorphic to , the cyclic group of order . 19. Find an integer and a field such that with , the cyclic group of order . Here is a primitive th root of unity over 8 . 20. Calculate the Galois group of the polynomial . 21. Let be transcendental over , prime. Show that the Galois group of is isomorphic to . More on Constructions
The following exercises show that not all regular gons can be constructed in the plane using only a straightedge and compass. The reader may refer to the exercises of Chapter 2 for the relevant definitions. Definition A complex number are both constructible. is constructible if its real and imaginary parts 22. Prove that the set of all constructible complex numbers forms a subfield of the complex numbers . 23. Prove that a complex number is constructible if and only if the real number and the angle (that is, the real number cos are constructible. 24. Prove that if is constructible, then both square roots of are constructible. Hint: use the previous exercise. 25. Prove that a complex number is constructible if and only if there exists a tower of fields , each one a quadratic extension of the previous one, such that . 26. Prove that if is constructible, then must be a power of . 27. Show that the constructibility of a regular gon is equivalent to the constructibility of a primitive th root of unity . Since the cyclotomic polynomial is irreducible over the rationals, we have deg . 28. Prove that is a power of if and only if has the form where are distinct Fermat primes, that is, primes of the form for some nonnegative integer . Hint: if is prime then must be a power of . Conclude that if does not have this form, then a regular gon is not constructible. For instance, we cannot construct a regular gon for , or . Gauss proved that if has the above form, then a regular gon can be constructed. See Hadlock (1978). Chapter 12 Cyclic Extensions Continuing our discussion of binomials begun in the previous chapter, we will , then show that if is a splitting field for the binomial where is a primitive th root of unity. In the tower the first step is a cyclotomic extension, which, as we have seen, is abelian and may be cyclic. In this chapter, we will see that the second step is cyclic of degree and can be chosen so that min . Nevertheless, as we will see in the next chapter, the Galois group need not even be abelian. In this chapter, we will also characterize cyclic extensions of degree relatively prime to expchar , as well as extensions of degree , but we will not discuss extensions of degree for , since this case is not needed and is considerably more complex. 12.1 Cyclic Extensions
Let be a field with expchar field for the binomial over distinct roots in . If is a root of then the roots of in are , let , where and let be a splitting . Note that has and is a primitive th root of unity over (12.1.1) and so . In words, all th roots of can be obtained by first adjoining the th roots of unity and then adjoining any single th root of . 262 Field Theory The extension can thus be decomposed into a tower where the first step is cyclotomic. For the second step, it will simplify the notation to simply assume that Hence, . is finite, Galois and the base field As to the Galois group value on and of contains all the th roots of unity. , each is uniquely determined by its for some . In fact, the map is an embedding of into , and so is isomorphic to a subgroup of and is therefore cyclic of degree . This follows easily from the assumption that contains the th roots of unity, for if , , then and so Definition Let expchar if is a root of a binomial Note that if type . . . An extension over , that is, if and if , then is pure of type . is also pure of is pure of type We can now provide a characterization of cyclic extensions when the base field contains the th roots of unity. Theorem 12.1.1 Let expchar . Suppose that contains the roots of unity and let . Then the following are equivalent: 1) is pure of type . 2) is cyclic of degree . In this case, is a root of for some if and only if min for some . Proof. We have seen that a pure extension of type is cyclic of type . For the converse, assume that is cyclic of degree , with Galois group th Cyclic Extensions 263 We are looking for a field primitive element that is a root of a binomial of the form , for . The roots of any polynomial have the form and the roots of the binomial have the form where which is a primitive th root of unity. Hence, if we can find an , then min for Since the product of these roots is in , we have and so min type , and therefore also of type . . Hence, is pure of Thus, we are left with finding an for which is a primitive th root of unity. This is the content of Hilbert's Theorem 90, which we prove next. We leave proof of the final statement of this theorem as an exercise. Theorem 12.1.2 (Hilbert's Theorem 90) Let be a finite cyclic extension of degree , with Galois group . An element has the form for some nonzero if and only if In particular, if the base field contains a primitive th root of unity and the previous statement applies. Proof. If , then and so , then For the converse, suppose that . We seek an element for which , that is, an element fixed by the operator , where is multiplication by . This suggests looking at the elements 264 Field Theory for , which have the property that for . Hence, the sum is a promising candidate, since applying shifts each term to the next, except for the last term. But since , we have and so applying desired. wraps the last term to the first. Hence, , as However, there is a problem. We do not know that change in the definition of is in order. Let is nonzero. Accordingly, a with , where is an as yet undetermined element of previous analysis still applies. In particular, and if . Then the then since we again have . But now, since the automorphisms are distinct, the Dedekind independence theorem implies that the linear combination is nonzero, and so there must be a nonzero This is our . For the last statement, if , we have for which is nonzero. is a primitive th root of unity, then since Cyclic Extensions 265 and the previous statement applies. 12.2 Extensions of Degree Char
There is an "additive" version of Theorem 12.1.1 that deals with cyclic extensions of degree equal to char , where the role of the binomial is played by the polynomial . Suppose that is a field of characteristic is a root of the polynomial . Let and suppose that for . Since for all , we have and so the distinct roots of are Hence, is a splitting field of . (In contrast to the previous case, we need no special conditions on , such as containing roots of unity, to ensure that if an extension of contains one root of , it contains all the roots of .) If then , with roots splits in . If , then min has degree where . The sum of these roots is , for some integer , and since this number lies in , but since , it follows that , whence min is irreducible. In short, either splits in or is irreducible over with splitting field , for any root of . Since is a splitting field for the separable polynomial , it follows that is Galois. If is irreducible over and , there exists a for which . Since , it follows that is the cyclic group generated by . Definition An extension of degree if is a root of an irreducible binomial char over is pure of type . 266 Field Theory Theorem 12.2.1 (ArtinSchreier) Let char . The polynomial either splits in or is irreducible over . An extension is cyclic of degree if and only if it is pure of type . Proof. We have seen that an extension that is pure of type is cyclic of type . Suppose that is cyclic of degree , with Galois group . If has the property that , then the roots of min are Moreover, since it follows that and so min . To find such an element, we need the additive version of Hilbert's Theorem 90. Theorem 12.2.2 (Hilbert's Theorem 90, Additive Version) Let be a finite cyclic extension with Galois group . An element has the form for some if and only if Tr . Proof. Assume that Tr . Let and consider the map It is easy to verify that for then Tr and so if Tr Thus, is the desired element. (Since the trace map is the sum of the automorphims in the Galois group, it is not the zero map and so there is a for which Tr .) Proof of the converse is left to the reader. In this section and the previous one, we have discussed cyclic extensions of degree where expchar or char . A discussion of cyclic extensions of degree for is quite a bit more involved and falls beyond the intended scope of this book. The interested reader may wish to consult the books by Karpilovsky (1989) and Lang (1993). Exercises
1. Assume that contains the th roots of unity and suppose that . Show that is a root of a binomial over if and only if it is a root of an irreducible binomial over , where . Cyclic Extensions 267 2. 3. Let that if Tr Let 4. 5. 6. 7. 8. 9. be a finite cyclic extension, with Galois group . Show has the form for some , then . be cyclic of degree where is a prime. Let with cyclic of degree where . Let and suppose that . Show that . Let char and let be cyclic of degree over , where min . Show that where and . Let be a field and let be the extension of generated by the th roots of unity, for all . Show that is abelian. Let be a field and let be an automorphism of of order . Suppose that has the property that and . Prove that there exists an such that . Let be a field and let be an automorphism of of order . Show that there exists an such that . Let be finite and abelian. Show that is the composite of fields such that is cyclic of primepower degree. Thus, the study of finite abelian extensions reduces to the study of cyclic extensions of primepower degree. Let be a field containing the th roots of unity. We do not assume that expchar . Let be an algebraic closure of . Show that if is separable over and if is a root of the binomial with , then is cyclic of degree . Chapter 13 Solvable Extensions We now turn to the question of when an arbitrary polynomial equation is solvable by radicals. Loosely speaking, this means (for char ) that we can reach the roots of by a finite process of adjoining th roots of existing elements, that is, by a finite process of passing from a field to a field , where is a root of a binomial , with . We begin with some basic facts about solvable groups. 13.1 Solvable Groups
Definition A normal series in a group is a tower of subgroups where . A normal series is abelian if each factor group abelian, and cyclic if each factor group is cyclic. is Definition A group is solvable (or soluble) if it has an abelian normal series. Theorem 13.1.1 The following are equivalent for a nontrivial finite group . 1) has an abelian normal series. 2) has a cyclic normal series. 3) has a cyclic normal series in which each factor group is cyclic of prime order. Proof. It is clear that 3) 2) 1). Thus, we need to prove only that 1) 3). Let be an abelian normal series. We wish to refine this series by inserting subgroups until all quotients have prime order. The Correspondence Theorem says that the natural projection is a normalitypreserving bijection from the subgroups of containing to the subgroups of . Hence, by Cauchy's Theorem, if a prime divides then has a subgroup of order , which must have the form for . 270 Field Theory is abelian, . Finally, since Theorem implies that Since and so . Thus, is abelian, the Third Isomorphism is also abelian. Thus, we have refined the original abelian normal series by introducing , where has prime order. Since is a finite group, we may continue the refinement process until we have an abelian normal series, each of whose quotient groups has prime order. The next theorem gives some basic properties of solvable groups. The proofs of these statements, with the possible exception of 2), can be found in most standard texts on group theory. Theorem 13.1.2 1) Any abelian group is solvable. 2) (FeitThompson) Any finite group of odd order is solvable. 3) (Subgroups) Any subgroup of a solvable group is solvable. 4) (Quotients) If is solvable and , then is solvable. 5) (Lifting property) If then and solvable imply that is solvable. 6) (Finite direct products) The direct product of a finite number of solvable groups is solvable. 7) The symmetric group is solvable if and only if . 13.2 Solvable Extensions
Although our results can be proved in the context of arbitrary finite extensions, we shall restrict our attention to separable extensions. As the reader knows, this produces no loss of generality for fields of characteristic or finite fields. Moreover, if is an inseparable polynomial, then where is separable. Thus, with respect to the solution of polynomial equations, the restriction to separable extensions is not as severe as it might first appear. Definition A finite separable extension is solvable if the finite Galois extension nc has solvable Galois group, where nc is the normal closure of over . Theorem 13.2.2 The class of solvable extensions is distinguished. Solvable Extensions 271 Proof. Speaking in general, consider a finite separable tower of the form Since the first step is normal, we have Now, if the Galois groups and of each step are solvable, then the quotient is solvable as well. Hence, by the lifting property of solvability, the Galois group of the full extension is solvable. On the other hand, if the full extension has solvable Galois group , then the Galois group of the lower step, being isomorphic to a quotient of is solvable and the Galois group of the upper step, being a subgroup of is solvable. Thus, in such a tower, solvability of the Galois groups has the "tower property." Note also that the implication solvable implies solvable does not require that the lower step be normal. Now we can get to the business at hand. Suppose first that where the full extension separable, and we have the tower nc is solvable. The lower step is finite and nc where the full extension has solvable Galois group, and therefore so does the lower step. That is, is solvable. As to the upper step , it is finite and separable. Consider the tower nc nc Since the full extension has solvable Galois group, so does the full extension nc nc (which is an upper step of the previous tower). Hence, the lower step has solvable Galois group, that is, is solvable. We have shown that if the full extension is solvable, so are the steps. Suppose now that each step and solvable and consider Figure 13.2.1. Since all extensions are finite and separable, we will have no trouble there. 272 Field Theory nc(N/F) N=nc(K/F)nc(E/K) nc(K/F) E K F nc(E/K) Figure 13.2.1 Since nc it is sufficient to show that nc nc is solvable. nc is isomorphic to a To this end, Theorem 6.5.6 implies that subgroup of the finite direct product hom Since this is a finite direct product and since each conjugate is isomorphic to , it suffices to show that is solvable. Consider the tower nc The group nc is solvable since is solvable. As to the Galois group nc of the upper step, it is a subgroup of the Galois group . Thus, it is sufficient to show that is solvable. But is the Galois group of a composite and is therefore isomorphic to a subgroup of the nc and nc , both of direct product of the Galois groups which are solvable, precisely because the lower and upper steps in the tower are solvable. For the lifting property, if is solvable and nc Lifting gives nc and the full extension is finite and Galois, and nc
nc is arbitrary, then nc nc and since the latter is solvable, so is the former. Solvable Extensions 273 13.3 Radical Extensions
Loosely speaking, when char , an extension is solvable by radicals if it is possible to reach from by adjoining a finite sequence of th roots of existing elements. More specifically, we have the following definitions, which also deal with the case char . Definition Let expchar tower of fields and let . A radical series for is a such that each step Pure of Class 1 is one of the following types: where is an th root of unity, where we may assume without loss of generality that . Pure of Class 2 where is a root of , with and . Pure of Class 3 (For only) where is a root of , with . For steps of classes 1 and 2, the number is the exponent (or type) of the step. The exponent of a class 3 step is . A finite separable extension that has a radical series is called a radical extension. If char , we may assume that the exponent in a class 1 extension is relatively prime to , for if is an th root of unity where and , then is also an th root of unity. Note that lifting a radical series gives another radical series with the same class of steps, for if , where is a root of , then where is a root of . For convenience, we write or to denote the fact that extension . is a radical series for the 274 Field Theory Theorem 13.3.1 (Properties of radical extensions) 1) (Lifting) If is a radical extension and , then the lifting is a radical extension. 2) (Each step implies full extension) If , where and are radical extensions, then so is the full extension . 3) (Composite) If and are radical extensions, then so is the composite extension . 4) (Normal closure) If is a radical extension, then so is nc . Proof. For 1), let . Lifting the series by gives the radical series and so For part 2), if is a radical extension. and , then lift the series by : and append it to the end of to get and so For part 3), if of is a radical extension. and by are radical extensions, then so is the lifting and so is the full extension . For part 4), the normal closure is nc
hom Since is a finite separable extension, hom is a finite set. Hence, the composite above is a finite one. We leave it as an exercise to show that if , then . Hence, is a radical extension, and therefore so is the finite composite nc . 13.4 Solvability by Radicals
We are interested in extensions extension . where is contained in a radical Definition A finite separable extension is solvable by radicals if , where is a radical extension. Theorem 13.4.1 1) The class of extensions that are solvable by radicals is distinguished. Solvable Extensions 275 2) If is solvable by radicals then so is nc . In fact, if where is a radical extension, then nc nc where Proof. Let nc is a normal radical extension. If is solvable by radicals then with radical. Hence, the lower step the upper step, radical implies solvable by radicals. Now suppose the steps in the tower are radical. is solvable by radicals. For is radical and so is KRK/F RE/K E K F
Figure 13.4.1 Referring to Figure 13.4.1, we have RK/F and where and are radical extensions. Lifting by gives the radical extension and so the tower is radical. It follows that the full extension is radical and so by radicals. is solvable 276 Field Theory As to lifting, if with radical, the lifting by gives and since is radical, is solvable by radicals. with The second part of the theorem follows from the fact that if is radical, then nc with nc radical. nc 13.5 Solvable Equivalent to Solvable by Radicals
Now we come to the key result that links the concepts of solvable extension and solvability by radicals. Here we employ the results of Chapter 12 on cyclic extensions, taking advantage of the fact that we may assume that all appropriate roots of unity are present. Theorem 13.5.1 A finite separable extension is solvable by radicals if and only if it is solvable. Proof. Suppose that is solvable. We wish to show that is solvable by radicals. By definition, nc is solvable and if we show that nc is also solvable by radicals, then the lower step is also solvable by radicals. Thus, we may assume that is normal. As to the presence of roots of unity, let . If does not contain a primitive th root of unity , then we can lift the extension by adjoining to get which is also solvable and normal. If we show that this extension is solvable by radicals, then so is the tower since the lower step Hence, the lower step Note that since is solvable by radicals (being pure of class 1). is also solvable by radicals. is finite and Galois, Corollary 6.5.3 implies that Hence, contains a primitive th root of unity, where . Thus, we may assume that is normal and contains a primitive th root of unity, where . It follows that if is any prime dividing , then contains a primitive th root of unity. Solvable Extensions 277 Since is finite, Galois and series decomposition is solvable, there is a normal (13.5.1) where and is cyclic of prime order . Taking fixed fields gives a tower fix dividing (13.5.2) Let us examine a typical step in this series. The relevant piece of the Galois correspondence is shown in Figure 13.5.1. E Fi Fi+1
Figure 13.5.1 Gi+1 Gi {1} Since is finite and Galois, the Galois correspondence is completely closed, that is, all intermediate fields and subgroups are closed. Thus, since is normal in , it follows that is normal (and hence Galois) and that which is cyclic of degree . Hence, base field contains the th roots of unity. is a cyclic extension whose Now, if char , then Theorem 12.2.1 implies that is pure of class 3. On the other hand, if char , then expchar and Theorem 12.1.1 implies that is pure of class 1 or class 2. Thus, is solvable by radicals, as desired. For the converse, suppose that is solvable by radicals, with where is a radical extension. Then Theorem 13.4.1 implies that the full extension in the tower nc has a radical series nc Let be the product of the types of all the steps in this series. Lift the tower by nc 278 Field Theory adjoining a primitive th root of unity , to get the radical series nc . Note that if is of class 1, then the step is trivial, and we may remove it. Thus, we may assume that all steps in the lifted tower are pure of class 2 or class 3. It follows from Theorems 12.1.1 and 12.2.1 that these pure steps are cyclic and so Theorem 6.6.2 implies that the Galois group is solvable. nc We have seen in the proof of 13.2.2 that since nc nc nc which contains where the full extension has solvable Galois group, so does the lower step. Hence, nc is a radical extension, which implies that is solvable by radicals. 13.6 Natural and Accessory Irrationalities
Let us assume that char solvable by radicals. Let of unity. and suppose that and assume that is finite, normal and contains the th roots Then, by definition, there is a radical series of the form (13.6.1) where . A typical step in this series has the form , where . Elements of the form , for , might reasonably be referred to as irrationalities, at least with respect to (or ). Kronecker coined the term natural irrationalities for those irrationalities of that lie in and accessory irrationalities for those irrationalities of that do not lie in . Given a radical series (13.6.1) containing there is another radical series , it is natural to wonder whether that contains only natural irrationalitites, that is, for which the top field itself. is We begin by refining the steps in (13.6.1) so that each has prime degree. Consider a step . Since the steps in the series are cyclic, every subgroup of the Galois group of is normal, and so all lower steps are normal. If where is prime, then has a subgroup of index and so fix is cyclic of degree . Solvable Extensions 279 Hence, any step of (13.6.1) can be decomposed into a tower , where the lower step is cyclic of prime degree. The upper step has the form and is cyclic of degree . We may repeat this decomposition on the upper step until each step is decomposed into a tower of cyclic extensions of prime degree. So, let us assume that each step in (13.6.1) is cyclic of prime degree. Consider the tower obtained by intersecting each field in (13.6.1) by (13.6.2) We wish to show that each step in (13.6.2) is also cyclic of prime degree. This is the content of the following theorem. It will follow that has a radical series that starts at and ends precisely at . Theorem 13.6.1 Let char . Let , where prime degree . Let be finite and normal. Then either trivial or Galois of degree . Proof. Figure 13.6.1 shows the situation.
B=A( )
normal prime deg is Galois of is E E=(A E)( ) E A Figure 13.6.1 We first show that Galois correspondence of fix (the reverse inclusion is clear). The plan is as follows. Let fix Since follow that is normal, if we show that fix and so for any , as desired. implies , it will is normal by showing that it is closed in the , that is, But it is sufficient to show that since then , To this end, since ) is finite and separable, it is simple, say 280 Field Theory where . If , then the extension is trivial, so assume need to show that if then , since then . In the tower . We the entire extension has prime degree . Since over and the lower step is nontrivial. Hence, is normal, the minimal polynomial and so its roots lie in min min min splits . Also, since each is a root of min and since is normal (being an upper step of , it follows that for all . Hence, But each sends to a conjugate is what we needed to prove and shows that Finally, we must show that is the lifting of by Galois, the Galois group of the lifting satisfies of , and so is normal. . To see this, note that . Since . This is and so . Theorem 13.6.2 (The theorem on natural irrationalities) Let char . Let be finite and normal. Let and assume that contains the th roots of unity. If is solvable by radicals, then there is a radical series starting with and ending with . We remark that the requirement that contain the appropriate roots of unity is necessary. An example is given by the casus irreducibilis, desscribed in the exercises. 13.7 Polynomial Equations
The initial motivating force behind Galois theory was the solution of polynomial equations . Perhaps the crowning achievement of Galois theory is the statement, often phrased as follows: There is no formula, similar to the quadratic Solvable Extensions 281 formula, involving only the four basic arithmetic operations and the taking of roots, for solving polynomial equations of degree or greater over . However, this is not the whole story. The fact is that for some polynomial equations there is a formula and for others there is not, and, moreover, we can tell by looking at the Galois group of the polynomial whether or not there is such a formula. In fact, there are even algorithms for solving polynomial equations when they are "solvable," but these algorithms are unfortunately not practical. Let us restrict attention to fields of characteristic . We refer to the four basic arithmetic operations (addition, subtraction, multiplication and division) and the taking of th roots as the five basic operations. Let be a field of characteristic . We will say that an element is obtainable by formula from if we can obtain by applying a finite sequence of any of the five basic operations, to a finite set of elements from . If is a pure extension, it is clear that any element of , being a polynomial in , is obtainable by formula from . Hence, any element of a radical extension is obtainable by formula from . Conversely, if is obtainable by formula from , then there is a finite set and a finite algorithm for obtaining from , where each step in the algorithm is the application of one of the five basic operations to elements of some extension of . If the operation is one of the four basic operations, then the result of the application is another element of the field . It the operation is the taking of a root, then the result will lie in a pure extension of . Thus, all the operations in the algorithm can be performed within a radical extension of . Hence, lies in a radical extension of . Theorem 13.7.1 Let be a field of characteristic . An element can be obtained by formula from if and only if lies in a radical extension of , that is, if and only if is solvable by radicals. Let us say that a root of a polynomial over is obtainable by formula if we can obtain by formula from . Thus, a root of is obtainable by formula if and only if is solvable by radicals. Theorems 13.4.1 and 13.5.1 now imply the following. Theorem 13.7.2 Let char polynomial over . Let over . and let and let be a be a splitting field for 282 Field Theory 1) The roots of are obtainable by formula if and only if the extension is solvable. 2) Let be irreducible over . One root of is obtainable by formula if and only if all roots of are obtainable by formula. According to Theorem 11.5.7, for any prime number , there exists an irreducible polynomial of degree over whose Galois group is isomorphic to . Since the group is not solvable for , Theorem 13.7.2 implies that if 5, then none of the roots of can be obtained by formula. Although it is much harder to show, this also holds for any positive integer see Hadlock, 1987 . Thus, we have the following. Theorem 13.7.3 For any , there is an irreducible polynomial of degree over , none of whose roots are obtainable by formula. As a consequence, for any given , there is no formula for the roots, similar to the quadratic formula, involving only the four basic operations and the taking of roots, that applies to all polynomials of degree . More specifically, we have Corollary 13.7.4 Let and consider the generic polynomial , where are algebraically independent over . Then there is no algebraic formula, involving only the five basic operations, the elements of and the variables , with the property that for any polynomial of degree over , we can get a root of by replacing in the formula by , for all . Exercises
1. 2. Prove that if then is solvable if and only if and are solvable. Prove that if is a radical series, then there is a radical series that is a refinement of this series (formed by inserting additional intermediate fields) for which each extension has prime exponent. Prove that if is solvable by radicals and hom then is also solvable by radicals. Calculate the Galois group of the polynomial . Is there a formula for the roots? Prove that if is a polynomial of degree over with Galois group isomorphic to then is irreducible and separable over . While the class of (finite, separable) solvable extensions is distinguished, show that the class of Galois solvable extensions does not have the tower property, and so is not distinguished. Hint: use the FeitThompson result (Theorem 13.1.2) and the proof of Theorem 11.5.1. Prove that a finite separable extension of characteristic is solvable by radicals if and only if there exists a finite extension with 3. 4. 5. 6. 7. Solvable Extensions 283 8. and a radical series for in which each step is one of the following classes: 1) where is an th root of unity with prime and . 2) where is a root of , with , prime and . 3) where is a root of the irreducible polynomial , with . Prove Theorem 13.7.2. Hint: for part 2), consider the normal closure of , where is an obtainable root of . Casus Irreducibilis
Cardano's formula for the cubic equation is This formula does not always yield a "satisfactory" solution, especially to the interested parties of the 16th century. For instance, the equation has only one real solution , but Cardano's formula gives (which must therefore equal , a handy formula to remember). The most serious "problem" with Cardano's formula comes when since in this case, the formula contains the square root of a negative number, something Cardano referred to as "impossible", "useless" and whose manipulation required "mental torture". For instance, the equation has a simple real solution , but Cardano's formula gives (which is equal to ). Cases where are known as casus irreducibilis and were the subject of much debate in the 1500s. Efforts to modify the formula for the solution of a cubic with three real roots in order to avoid nonreal numbers were not successful, and we can now show why. (Actually, this turned out to be a good thing, since it sparked the development of the complex numbers.) 9. Let be an irreducible cubic over with three real roots. This exercise shows that no root of can be obtained by formula if we allow the 284 Field Theory taking of real th roots only, that is, we show that no root of contained in a radical series that is completely contained in . a) Suppose that a root of is contained in a radical series is Let be the discriminant and let series . Show that there is a radical containing with . b) Show that the radical series in part a) can be refined (by inserting more intermediate fields) into a radical series in which each step has prime exponent. c) Let be the first index such that a root of is in and consider the extension . Show that is a splitting field for over . d) Since is pure of prime exponent, we have , where is a root of , with prime and . More generally, prove that if is a field, and is a prime, then the polynomial is either irreducible over or . Hint: Suppose that where deg and deg . If are the roots of and are the roots of , then Take the th power of this and use . Then use the fact that . e) Show that is not possible. (The primitive th roots of unity do not lie in .) Hence, is irreducible. f) Show that and . g) Show that is normal. What does that say about the roots of ? Galois' Result
Galois, in his memoir of 1831, proved the following result (Proposition VIII): "For an equation of prime degree, which has no commensurable divisors, to be solvable by radicals, it is necessary and sufficient that all roots be rational functions of any two of them." In more modern language, this theorem says that if is irreducible and separable of prime degree , then the equation is solvable by radicals if and only if is a splitting field for , for any two roots and of Solvable Extensions 285 . To prove this theorem, we require some results concerning solvable transitive subgroups of , the group of permutations of . Any map defined by , where with is called an affine transformation of . Let aff be the group of all affine transformations of . Note that aff is a subgroup of . The translations are the affine maps . Let trans be the subgroup of aff consisting of the translations. Let be translation by . Two elements and of are conjugate if there is a for which . 10. a) Show that trans is a normal subgroup of aff . b) Show that is the cycle , that any nonidentity translation is a cycle and that an element is a cycle if and only if it is conjugate to . c) Within aff , the nonidentity translations are characterized as having no fixed points, whereas all elements of aff trans have exactly one fixed point. d) Show that aff acts transitively on . e) Show that aff and trans . Hence, trans is a Sylow subgroup of aff and is the only subgroup of aff of order . f) Show that trans , aff trans and aff are solvable. 11. Prove that if has the property that aff , then aff . 12. The following are equivalent for a subgroup of : 1) is transitive. 2) contains a subgroup conjugate to trans , that is, , for some cycle . 13. The following are equivalent for a transitive subgroup of : 1) The only element of with two fixed points is the identity. 2) is conjugate to a subgroup of aff . We have proved that for a transitive subgroup of , the first two statements below are equivalent. We now add a third. 1) The only element of with two fixed points is the identity. 2) is conjugate to a subgroup of aff . 3) is solvable. It is clear that 2) implies 3), since a conjugate of a solvable group is solvable. The next few exercises prove that 3) implies 1). 14. If is a transitive subgroup of , show that any normal subgroup of also acts transitively on . 286 Field Theory 15. Let be a transitive, solvable subgroup of with prime indices . Then has a normal series a) Show that , where is a cycle. b) Show that the only element of that has two fixed points is the identity. We can now return to Galois' result concerning solvability by radicals for a primedegree equation . 16. Prove that if is irreducible and separable of prime degree , then the equation is solvable by radicals if and only if is a splitting field for , for any two roots and of . Part IIIThe Theory of Binomials Chapter 14 Binomials We continue our study of binomials by determining conditions that characterize irreducibility and describing the Galois group of a binomial in terms of matrices over . We then consider an application of binomials to determining the irrationality of linear combinations of radicals. Specifically, we prove that if are distinct prime numbers, then the degree of over is as large as possible, namely, products of the form . This implies that the set of all where numbers , is linearly independent over and . For instance, the are of this form, where , . Hence, any expression of the form where , must be irrational, unless for all . First, a bit of notation. If , then stands for a particular (fixed) root of . The set of primitive th roots of unity is denoted by and always denotes a primitive th root of unity. 14.1 Irreducibility
Let us first recall a few facts about the norm. Let If the minimal polynomial of min be finite with . 290 Field Theory has roots then where we have . Note that . Also, for all and , 1) The norm is multiplicative, that is, for all , In particular, for any positive integer . Also, 2) For , . and so 3) If are finite and if then Our technique for determining the irreducibility of a binomial for is an inductive one, beginning with the case prime. Theorem 14.1.1 Let be a prime. Then the following are equivalent: 1) 2) has no roots in 3) is irreducible over Proof. It is easy to see that 1) and 2) are equivalent and that 3) implies 2). To see that 1) implies 3), let be a root of in and assume that . We wish to show that , which implies that is min and is therefore irreducible. Since , taking the norm gives where for which . Now, if . Hence then and there exist integers and which is a contradiction. Thus , as desired. Binomials 291 To generalize this to nonprime exponents, assume that is a product of not necessarily distinct odd primes. (We will consider the even prime later.) Let us write as and where . Let be a root of in and write . Then Hence, is a root of and is a root of and we have the tower Repeating the process with , if , then so is a root of and is a root of over and we have the tower Clearly, we can repeat this process as desired to obtain a tower where , and where each step has the property that is a root of the binomial of prime degree over . Now, the binomial is irreducible if and only if and this happens if and only if each binomial is irreducible, which according to Theorem 14.1.1, is equivalent to the conditions (14.1.1) for all . Let us improve upon these conditions. 292 Field Theory First, note that if , that is, if for , then Hence, (14.1.1) is implied by the following conditions, which involve membership in a power of the base field only (14.1.2) for . Thus, under these conditions, each binomial irreducible, with root , and so is for char . Assuming that all the primes , this can be written as are odd or that (14.1.3) for . For , this is For , we get and applying the norm and using the case gives In general, if , then applying the norm to (14.1.3) gives Thus, for and we can rephrase the conditions (14.1.2) as for . Binomials 293 Theorem 14.1.2 If primes, then the binomial is a product of not necessarily distinct odd is irreducible if for . and write Let us now turn to the case where for . As seems often the case, the even prime problems. To illustrate, if , then for any causes additional and so the binomial is reducible even though . Thus, for , we must at least include the restriction that for any , that is, that . It turns out that no further restrictions are needed. Theorem 14.1.3 Let be a field and . 1) is irreducible if and only if 2) For , the binomial is irreducible if and only if and . Proof. Part 1) is clear. For part 2), assume that is irreducible, where . If for some , then is reducible. Hence, . Also, if for , then factors as above. Hence, and . and imply that For the converse, we show that the conditions is irreducible for all , by induction on . We have seen that this holds for . Assume that it holds for all positive integers less than . Let be a root of in a splitting field and write Hence, is a root of and is a root of and we have the tower 294 Field Theory The lower step has degree since . As to the upper step, if and , then the induction hypothesis implies that is irreducible over , in which case which implies that 2) holds. Hence, we need only consider the two cases wherein these hypotheses fail. , that is, if , for some , we claim that as well. The problem is that may not be a square in . But taking norms gives If where . Hence, . It follows that So, if either condition fails, then We must show that this implies that Applying the norm gives for some . . is irreducible over where is a root of . Since in , then , it follows that . In other words, if . Over , we have the factorization (14.1.2) If both of the factors on the right side are irreducible over , then cannot factor nontrivially over , because each irreducible factor of over , being over as well, would be a multiple of one of the irreducible factors on the right of (14.1.2) and so would have degree greater than . But two such factors would then have product of degree greater than . Thus, in this case, is irreducible over . On the other hand, if one of the factors in (14.1.2) is reducible, the induction hypothesis implies that one of or is in either or . Thus, in either case, one of or is in , say Thus, and . It follows that , a contradiction to the hypothesis of the lemma. Hence, this case does not occur. Now we can prove the main result of this section. Binomials 295 Theorem 14.1.4 Let be an integer and let . 1) If , then is irreducible over if and only if for all primes . 2) If , then is irreducible over if and only if for all primes and . Proof. Assume first that is irreducible. Then for any prime , the polynomial is irreducible, for if is a nontrivial factorization, then is a nontrivial factorization of . Hence, by Theorem 14.1.3, for any . Also, if then the polynomial is irreducible and so again by Theorem 14.1.3, . Alternatively, we have a direct factoization For the converse, assume that for all primes and that when , we also have . We proceed by induction on . If , the result follows from Theorem 14.1.3. Assume that the theorem is true for integers greater than and less than . If , where , then Theorem 14.1.3 applies. Otherwise, has an odd prime factor . Suppose that where and . Let be a root of is a root of . Then is a root of and The induction hypothesis implies that first step in the tower is irreducible over and so the has degree . If degree , whence irreducible. is irreducible over , then the second step will have and , which is min We apply the inductive hypothesis to show that odd, we need only show that . If taking norms gives is irreducible. Since is for some then If is odd, we get is odd, we have is irreducible over , a contradiction. If is even then since , again a contradiction. Hence, and is irreducible over . 296 Field Theory 14.2 The Galois Group of a Binomial
Let us now examine the Galois group of a binomial and relatively prime to expchar . If is a root of all the roots are given by over and , for , then and so tower is a splitting field for over . Moreover, in the (14.2.1) the first step is a cyclotomic extension, which is abelian since its Galois group is isomorphic to a subgroup of . The second step is pure of type and so, according to Theorem 12.1.1, it is cyclic of degree and min Despite the abelian nature of the lower step and the cyclic nature of the upper step, the full extension (14.2.1) need not be abelian. The fact that and both satisfy simple polynomials over is the key to describing the Galois group . Since any must permute the roots of , there exists an integer for which Moreover, since is a normal extension of , the restriction of to is in and therefore sends to another primitive th root of unity, that is, where Multiplication in . has the following form. For , and There is something reminiscent of matrix multiplication in this. Indeed, let be the set of all matrices of the form Since Binomials 297 we see that nonsingular the product is a subgroup of the general linear group GL of all matrices over . Comparing this product with the action of shows that the map defined by satisfies and is, in fact, a monomorphism from Since if and only if , where into . is surjective is the Euler phi function, the map But in the tower we always have and . Hence is surjective (and an isomorphism) if and only if equality holds in these two inequalities. Theorem 14.2.1 Let be a positive integer relatively prime to expchar . Let be the splitting field for over , where . Let be a root of and . In the tower the first step is a cyclotomic extension and the second step is cyclic of degree with min . Also, is isomorphic to a subgroup of the group described above, via the embedding where and . The map is an isomorphism and if and only if both steps in the tower (14.2.1) have maximum degree, that is, if and only if 1) 2) , or equivalently, is irreducible over . 298 Field Theory A Closer Look
There are two issues we would like to address with regard to the previous theorem. First, statement 2) is phrased in terms of and we would prefer a statement involving only the base field . Second, we would like to find conditions under which is abelian. We will see that for an odd integer relatively prime to expchar , we can replace condition 2) with the condition that is irreducible over . With respect to the commutativity of , we will derive a general necessary and sufficient condition. However, we will first prove a simpler result; namely, assuming that , then is abelian if and only if the second step in (14.2.1) is trivial, that is, if and only if splits over . The Prime Case We first deal with both issues for prime. Recall that according to Theorem 14.1.1, the following are equivalent: 1) 2) has no roots in 3) is irreducible over Over the base field , which contains all the th roots of unity, we have or and the following are equivalent: 1) 2) 3) has no roots in 4) does not split over 5) is irreducible over The next lemma ties these two situations together, and strengthens statement 2) of Theorem 14.2.1 for prime. Lemma 14.2.2 Let be a prime and let . Then is irreducible over if and only if it is irreducible over . Proof. Certainly, if is irreducible over , it is also irreducible over . For the converse, consider the tower Since is irreducible over , we have On the other hand, the first step in the tower has degree at most and the second step is cyclic of degree , whence or . Hence , which implies that is irreducible min over . Binomials 299 As to the question of when the Galois group since both steps in the tower is abelian in the prime case, are abelian, if either step is trivial, then is abelian. Thus, if or if then is abelian. The converse is also true when is prime. Lemma 14.2.3 Let be a prime and let . Let be a splitting field for over . Then the Galois group is abelian if and only if at least one step in the tower (14.2.1) is trivial, that is, if and only if either or is reducible over . Proof. As mentioned, if one step is trivial then is abelian. Suppose now that and is irreducible over . Since , it has a conjugate that is also not in . Let be defined by . Since is irreducible over , for each , the map may be extended to a map defined by For and and , we have and these are distinct since is not abelian. The General Case Armed with the previous results for use the following fact. . Hence, and do not commute and prime, we consider the general case. We Suppose that splits over and has a nonabelian splitting field extension . Then if is abelian, cannot split in because otherwise, there would be a splitting field of satisfying . But abelian implies that the lower step is abelian and since all splitting fields for over are isomorphic, this contradicts the fact that is nonabelian. Theorem 14.2.4 Let be an odd positive integer relatively prime to expchar . Let and suppose that contains no th roots of unity other than . Let be any abelian extension. Then is irreducible over if and only if it is irreducible over . Proof. Clearly, if is irreducible over , it is also irreducible over the smaller field . Suppose that is irreducible over . Then for every 300 Field Theory prime , the polynomial is irreducible over and therefore also over , by Lemma 14.2.2. Now, if were reducible over , then it would have a root in and since is normal, would split over . is a root But since does not contain any primitive th roots of unity, then if of in a splitting field, the tower has nontrivial steps and so is nonabelian by Theorem 14.2.3. It follows from previous remarks that cannot split over the abelian extension . Hence, is irreducible over for all primes dividing and so is irreducible over . If , then cannot contain any primitive th roots of unity for any , and so it cannot contain any th roots of unity other than . Thus, since is an abelian extension, we may apply Theorem 14.2.4 to get the following strengthening of Theorem 14.2.1, for odd. Corollary 14.2.5 Referring to Theorem 14.2.1, let be an odd positive integer relatively prime to expchar . Then if and only if and is irreducible over . Since , we have the following corollary. and is an odd is irreducible over Corollary 14.2.6 Referring to Theorem 14.2.1, if positive integer then if and only if . Thus, when has the largest possible degree (which includes the important special case , we see that if and only if is irreducible over . We show next that is abelian if and only if splits over , or equivalently, has a root in . Note that for any positive integers and , we have This follows from the fact that since , all the roots of lie in is a root of . and since Theorem 14.2.7 Let be an odd positive integer relatively prime to expchar . Let be the splitting field for over , where . Suppose that where . Then the following are equivalent. 1) is abelian Binomials 301 2) has a root in 3) has a root in and therefore splits over Proof. Clearly, 2) 3) 1). We must show that 1) implies 2). Suppose that is abelian and let be the largest divisor of for which , that is, for some . The proof will be complete if we show that , since implies that is a root of in . If , let be a prime dividing and consider the tower Note that is irreducible over , for if not, then for some , whence , in contradiction to the definition of . Hence and since , we deduce that neither of the first steps is trivial. Hence, Lemma 14.2.3 implies that the Galois group is not abelian. But this is a contradiction to 1). In the exercises, we ask the reader to provide a simple example to show that Theorems 14.2.4 and 14.2.7 fail to hold when is even. More on When Is Abelian We conclude this section by generalizing the previous theorem, in order to characterize (for odd), with no restriction on the lower step, precisely when is abelian. The proof follows lines similar to that of Theorem 14.2.7, but is a bit more intricate and since it involves no new insights, the reader may wish to skip it on first reading. However, the result is of interest since it shows how the relationship between the th roots of unity and the ground field play a role in the commutativity of . We first need a result that is of interest in its own right. The proof is left as an exercise. Theorem 14.2.8 Let a and be irreducible over contains a primitive th root of unity. Then and splitting field over if and only if for some prime to . and suppose that have the same and relatively Note that if is a field and is the group of th roots of unity over is a (cyclic) subgroup of and so is , for some . , then Theorem 14.2.9 Let be an odd positive integer relatively prime to expchar . Let be the group of th roots of unity over and let . If is the splitting field for , where , then is abelian if and only if . Proof. Since is cyclic, it follows that if and only if . Suppose first that for some . Then 302 Field Theory for some integer . (More precisely, given any th root of and any th root of , there exists a such that this equation holds.) The field is cyclic over , since the latter contains a primitive th root of unity . Therefore, since the extensions and are both abelian, so is the extension Finally, since , it follows that is abelian. For the converse, assume that is abelian. Let be the largest positive integer such that , and , say for . We need to show that . Suppose to the contrary that and let be a prime number dividing . Let be the largest power of such that . (As an aside, the hypothesis that is odd and in Theorem 14.2.7 implies that , whence .) The first step is to show that the extension is abelian. It is clear that the notation is a bit unwieldy, so let us set and note that since and . To see that this extension is abelian, we embed it in an abelian extension. Since we have for some and so Now, since , there is a positive integer , it follows that for which , and since is a positive integer. Hence, is a root of contained in that lies in , that is, . Hence, all roots of are Putting the pieces together gives Binomials 303 Since the composite and are abelian (the latter by assumption), is abelian and therefore so is We now propose to arrive at a contradiction by considering the tower Note that , whence is irreducible over , since otherwise for some , in contradiction to the definition of . . Since is irreducible over , We first take the case we have , whence Since , it follows that and so the lower step is not trivial. However, since , the upper step in the tower is also not trivial. Hence, Lemma 14.2.3 implies that the Galois group is not abelian, the desired contradiction. Now assume that . With regard to the first step in the tower, since both divide but does not, it follows that . Since , the binomial is either irreducible over splits over . But is a root of this binomial that is not in and so irreducible over . Since the roots of are and and or is for each that . , there is a for which is not abelian, we shall need only . To show (the identity) and There are two possibilities for the second step in the tower. If is irreducible over then we can extend and to elements of by defining
, , , 304 Field Theory and
, , , Then
, , , and
, , , which are distinct since contradiction. If . Hence, is not abelian, a is reducible over then . Thus for some and so . Since and are both irreducible over , it follows that and , whence . Thus, and have the same splitting field over and Theorem 14.2.8 implies that for some . Taking th powers gives, since , for , which contradicts the definition of . Thus, proved. and the theorem is *14.3 The Independence of Irrational Numbers
A familiar argument (at least for ) shows that if is a prime number then and so . Our plan in this section is to extend this result to more than one prime and to th roots for . Since the case in which is even involves some rather intricate details that give no further insight into the issues involved, we will confine our attention to odd. (The case is straightforward and we invite the reader to supply a proof of Theorem 14.3.2 for this case.) If is rational, the notations and will denote the real positive th root of . The results of this section were first proved by Bescovitch 1940 , but the method of proof we employ follows more closely that of Richards 1974 . Lemma 14.3.1 Let terms, that is, where be a positive rational number, expressed in lowest and . If is an integer then Binomials 305 if and only if and for positive integers and In particular, if is a prime, then . Proof. One direction is quite obvious. Suppose that and are positive integers and . Then , it follows that , say , and , say , which implies that . It follows that whence and so and . where and since . Hence, and , Suppose now that is odd and is prime. Since for any prime , Theorem 14.1.4 (or Eisenstein's criterion) implies that is irreducible over and so . Let us generalize this to more than one prime. Theorem 14.3.2 Let Then be an integer and let be distinct primes. Proof. As mentioned earlier, we confine our proof to the case that Let . Since is odd. it is sufficient to show that which we shall do by induction on . Let be a prime. Since is irreducible over and contains no th roots of unity other than , Theorem 14.2.4 implies that is also irreducible over . Hence, and the theorem holds for . and let be a Now let us suppose that the theorem is true for the integer prime distinct from the distinct primes . Let and If is not irreducible over then there exists a prime such that . Thus, is a linear combination, over , of terms of the form 306 Field Theory where . There are two cases to consider. Case 1: If the linear combination involves only one term, then where and not all are . If , this can be written in the form This says that the radicand; call it , is a positive rational number and the polynomial has a root in . According to Theorem 14.2.7, must also have a root in , which is not possible since does not have the form , for relatively prime integers . Hence, this case cannot occur. Case 2: At least two terms in the linear combination are nonzero. It follows that one of the primes , which we may assume for convenience is , appears to different powers in at least two distinct terms. Collecting terms that involve like powers of gives
2 (14.3.1) 's are where nonzero. Now, since and where at least two of the is a Galois extension (this is why we adjoined in the first place), the inductive hypothesis implies that its Galois group has size . Since any must send roots of to other roots, it must send to for some choice of . Since there are such choices, all these choices must occur. Thus, there is a for which is the identity on and Since for some
2 , applying to (14.3.1) gives We now multiply (14.3.1) by and subtract the previous equation to get Binomials 307 where at least one of the coefficients is nonzero. This is a contradiction to the inductive hypothesis. We have therefore established that is irreducible over and the proof is complete. Exercises
1. Let be relatively prime to char . Show that the group 2. 3. 4. 5. 6. is generated by two elements and , where , and . What is ? (Van der Waerden) Let be relatively prime to char . Show that the Galois group of is isomorphic to a subgroup of the group of linear substitutions modulo , that is, maps on of the form where , . Let . Show that the following are equivalent: a) , prime implies b) , prime implies but where is the multiplicative order of in . Prove the following without using any of the results of Section 14.1. If and then is irreducible over if and only if and are irreducible over . Let char and let be cyclic of degree , with Galois group . If there exists a with Tr show that there exists an for which the polynomial is irreducible over . Let char and let where . Show that the Galois groups of and 7. 8. 9. are the same. Let be a positive integer relatively prime to expchar and let be a primitive th root of unity over . Let be the splitting field 2 for over , where , . If and if then is not abelian. Show that Theorem 14.2.4 and Theorem 14.2.7 fail to hold when is even. Hint: , where is a primitive th root of unity. Prove the following: Let be a monic irreducible polynomial of degree over , with constant term . Let be an integer for which , and , for all primes . Then the polynomial is also irreducible over . 308 Field Theory 10. Let be a primitive th root of unity over , odd, and let be a root of over . Then is the splitting field for . Assume that . In this exercise, we determine the largest abelian subextension of . a) If is a group, the subgroup generated by all commutators , for , , is called the commutator subgroup. Show that is the smallest subgroup of for which is abelian. b) If the commutator subgroup of a Galois group is closed, that is, if for some , then is the largest abelian extension of contained in . c) The commutator subgroup of is and if is defined as in Theorem 14.2.1, then where d) Prove that extension of 11. Prove that if . , and so contained in . are distinct primes then is the largest abelian by induction on . 12. Show that . 13. Let and be irreducible over and suppose that contains a primitive th root of unity. Then and have the same splitting field over if and only if for some and relatively prime to . Hint: if the splitting fields are the same, consider how the common Galois group acts on a root of each binomial. 14. Let be a finite Galois extension and let have degrees and over , respectively. Suppose that . a) Show that if is a conjugate of and is a conjugate of , then there is a such that and . Hence, the conjugates of are . b) Show that if the difference of two conjugates of is never equal to the difference of two conjugates of then . c) Let be a prime different from char . Let and be irreducible over , with roots and , respectively. Show that if then . Chapter 15 Families of Binomials In this chapter, we look briefly at families of binomials and their splitting fields and Galois groups. We have seen that when the base field contains a primitive th root of unity, cyclic extensions of degree correspond to splitting fields of a single binomial . More generally, we will see that abelian extensions correspond to splitting fields of families of binomials. We will also address the issue of when two families of binomials have the same splitting field. 15.1 The Splitting Field
Let be a field containing a primitive th root of unity and consider a family of binomials given by where the family is the set of constant terms. We will refer to . as the exponent of If is the splitting field for , then is the splitting field for the family . Since each extension is Galois, so is and Theorem 6.5.4 implies that is isomorphic to a subgroup of the product Since each is cyclic of degree dividing , the group is the direct product of cyclic groups of order dividing and so is abelian with exponent . An abelian extension whose Galois group has exponent will be referred to as an abelian extension with exponent . Thus, if contains a primitive th root of unity, the splitting field of any family of binomials over of exponent is an abelian extension of with exponent . Happily, the converse is also true. 310 Field Theory Suppose that is an abelian extension with exponent . Let be any field for which where is finite. Since is abelian, so is . In addition, is finite and has exponent . Since a finite abelian group is a direct product of cyclic subgroups, we have where each is cyclic with exponent and hence order . Corollary 6.5.5 implies that is a composite where is cyclic of order . Since contains the th roots of unity and is cyclic, Theorem 12.1.1 implies that is the splitting field for min where family . Hence is the splitting field over for the It follows that is the splitting field for the union intermediate fields . , taken over all finite Theorem 15.1.1 Let be a field containing a primitive th root of unity. An extension is abelian with exponent if and only if is the splitting field for a family of binomials over of exponent . Definition Let be a field containing a primitive th root of unity. An extension is a Kummer extension of exponent if is abelian and has exponent . Thus, according to Theorem 15.1.1, the Kummer extensions of of exponent are precisely the splitting fields over of families of binomials of exponent . 15.2 Dual Groups and Pairings
Before proceeding, we need a few concepts from group theory. If and are groups, we denote by hom the set of all group homomorphisms from to . Note that hom is a group under the product with identity being the constant map Lemma 15.2.1 1) If and for all . are abelian groups, then hom hom hom Families of Binomials 311 2) Let be the group of all th roots of unity over a field abelian group of exponent , then hom Proof. We leave it as an exercise to show that the map hom defined by hom hom . If is a finite is an isomorphism, proving part 1 . For part 2), since product of finite cyclic groups, part 1) implies that we hom when is cyclic. Suppose that has order , then hom maps into , we have can be written as the need only show that has order . If since for any , Hence, hom hom hom by setting homomorphism. Then . Suppose that and define , which is easily seen to define a group is a cyclic subgroup of hom size at most and so hom hom . of order . But hom is cyclic of order has , whence Definition If , and are abelian groups, a pairing of map that is a bihomomorphism, that is, 1) For each , the map defined by homomorphism. 2) For each , the map defined by homomorphism. into is a is a group is a group A pairing is the analogue of a bilinear map between vector spaces (and is sometimes referred to as a bilinear map). Note that for all and and . If and , we set , (We will write pairing is the set as and as .) The left kernel of a and the right kernel is defined similarly: 312 Field Theory It is easy to see that these kernels are normal subgroups of their respective parent groups. Note that for all if and only if , that is, if and only if , or equivalently, . Similar statements hold for the right kernel. Thus, we may define a pairing from to by and this pairing is nonsingular, that is, both the left and right kernels are trivial. Theorem 15.2.2 Let be a nonsingular pairing from abelian groups and into , the group of th roots of unity over a field . Then and both have exponent and 1) is isomorphic to a subgroup of hom 2) is isomorphic to a subgroup of hom Moreover, is finite if and only if is finite, in which case 3) hom and hom 4) , in particular, . Proof. First observe that if , then for all , and so , whence and has exponent . A similar statement holds for . Now consider the map hom defined by , where . Since the map is a group homomorphism from to hom . If is the constant homomorphism then for all , that is, , whence . Hence, the map is injective and 1) holds. Similarly, 2) holds. It follows from Lemma 15.2.1 that if hom The dual argument shows that and so that the monomorphism is an isomorphism. We can now return to binomials. . This also implies is finite, then 15.3 Kummer Theory
While each family of binomials gives rise to a unique Kummer extension, different families may produce the same extension, that is, different families may have the same splitting field. We seek a collection of families of binomials Families of Binomials 313 such that there is a onetoone correspondence between families in the collection and Kummer extensions. Let us phrase the problem a little differently, for which we require some notation. Recall that if , then by we mean a particular (fixed) root of . If , we let denote the set of all th roots of all elements of . Also, if and is a nonnegative integer then . Let be a field containing a primitive th root of unity. Of course, we may identify a family of binomials of a fixed exponent with the set of constant terms. (Since binomials with zero constant term are not very interesting, we exclude such binomials.) Moreover, the splitting field for is . In seeking a bijective correspondence between sets of constant terms (that is, families of binomials) and splitting fields , it is natural to restrict attention to maximal sets that generate the given splitting field. As we now show, if for some , then * where is the multiplicative subgroup of generated by and the th powers * of elements of . To see this, note that any element * of has the form for and . The th roots of have the form and since each of the factors in this product is in , so is the product. Hence, nothing new is added to the splitting field by increasing the set of * constants to , that is, Thus, for sets of constant terms, we may restrict attention to the lattice intermediate subgroups satisfying of all Indeed, we will show that the association onto the class of all Kummer extensions also obtain a description of the Galois group For root of , let , and let and so of of is a bijection from with exponent . We will in terms of . and be a Kummer extension with Galois group . If is a root of then is also a
, 314 Field Theory for some th root of unity . We claim that does not depend on , that is, is simply multiplication by an th root of unity. To see this, if is another root of , then where and so and so ; that is, depends only on . defined by , for any with It follows that the map is welldefined (does not depend on and we may write (15.2.1) without ambiguity. Moreover, the map is a pairing of a group bihomomorphism. Specifically, we have Also, into and , that is, . and The left kernel of this pairing is for all Also, since we are assuming that , for all fix fix It follows that the pairing
* * given by is nonsingular. We may thus apply Theorem 15.2.2. Theorem 15.3.1 Let be a field containing a primitive th root of unity and let , then be a subset of an extension of . If , where * Families of Binomials 315 the pairing
* given by
* is nonsingular. Also, 1) and 2) case * have exponent is finite if and only if
* * is finite, in which and hom
* The previous theorem not only describes the Galois group of a Kummer extension, but allows us to show that the map , from to , is a bijection. Theorem 15.3.2 Let be a field containing a primitive th root of unity. Let be the class of all Kummer extensions with exponent and let be the class of all subgroups of containing * . Then the map * is a bijection from onto , with inverse given by . Proof. To show that the map in question is injective, suppose that , with , . If , then and so there exists a * finite subset of for which . Let be the subgroup generated by and * . Then and Note that is finitely generated by is finite. Theorem 15.3.1 implies that over
* * and hence * Let us now adjoin . Let Then and be the subgroup generated by and . 316 Field Theory Another application of Theorem 15.3.1 gives
* * and since we get . It follows that and since was arbitrary, . A symmetric argument gives , whence . This proves that the map is injective. We have seen that any Kummer extension in is a splitting field extension for a family of binomials with exponent . If is the set of constant terms and if is the subgroup of generated by and * then and so the map is surjective. Let Then
* be a Kummer extension with exponent and let . * is a subgroup of containing , that is, . It is clear that . For the reverse inclusion, let . Then for some , which implies that is a root of and so . This shows that and so . Hence, * is the inverse map. Exercises
1. Referring to Lemma 15.2.1, show that the map hom defined by hom hom 2. 3. 4. 5. is an isomorphism. Let be a finite group and let hom if for all and Let be a finite abelian group with exponent for all hom then . Let be a proper subgroup of a finite abelian let . Then there exists hom . Let be a subgroup of a finite abelian group hom Show that hom hom . Show that otherwise. . If satisfies group of exponent such that of exponent . Let and but 6. Let be a subgroup of a finite abelian group hom of exponent . Let Show that hom . Families of Binomials 317 7. 8. Let be a family of binomials over of varying degrees. Suppose that for all and that contains a primitive th root of unity. Show that there is a family of binomials over , each of which has degree , with the same splitting field as . In this exercise, we develop the analogous theory for families of char polynomials of the form where . a) Prove that is abelian with exponent if and only if is the splitting field of a family of the form . b) Let be the map . Let such that . Let be the class of all additive subgroups of with . Let be the class of all abelian extensions of with exponent . Prove the following theorem: The map is a bijection between and . If is in and has Galois group then there is a welldefined pairing given by for any . The left kernel is and the right kernel is . The extension is finite if and only if is finite, in which case and . Appendix Mobius Inversion Mo bius inversion is a method for inverting certain types of sums. The classical form of Mo bius inversion was originally developed independently by P. Hall and L. Weisner, in 1935. However, in 1964, GianCarlo Rota generalized the classical form to apply to a much wider range of situations. To describe the concept in its fullest generality, we require some facts about partially ordered sets. Partially Ordered Sets
Definition A partial order on a nonempty set is a binary relation, denoted by and read "less than or equal to," with the following properties: 1) (reflexivity) For all , 2) (antisymmetry) For all and 3) (transitivity) For all , and , implies implies Definition A partially ordered set is a nonempty set , together with a partial order defined on . The expression is read " is less than or equal to ." If , we denote the fact that a is not less than or equal to by . Also, we denote the fact that , but , by . If there exists an element for which for all , we call a zero element and denote it by . Similarly, if there exists an element for which for all , then we call a one and denote it by . As is customary, when the partial order phrase "let be a partially ordered set." is understood, we will use the 320 Field Theory Note that in a partially ordered set, it is possible that not all elements are comparable. In other words, it is possible to have with the property that and . Thus, in general, is not equivalent to . A partially ordered set in which every pair of elements is comparable is called a totally ordered set or a linearly ordered set. Example A.2.1 1) The set of real numbers, with the usual binary relation , is a partially ordered set. It is also a totally ordered set. 2) The set of natural numbers, together with the binary relation of divides, is a partially ordered set. It is customary to write (rather than ) to indicate that divides . 3) Let be any set, and let be the power set of , that is, the set of all subsets of . Then , together with the subset relation , is a partially ordered set. z Definition Let is the set be a partially ordered set. For , the (closed) interval We say that the partially ordered set is a finite set. Notice that if is locally finite if every closed interval is locally finite and contains a zero element , then the set is finite for all , for it is the same as the interval . The Incidence Algebra of a Partially Ordered Set
Now let be a locally finite partially ordered set, and let : Addition and scalar multiplication are defined on 0 if by be a field. We set and We also define multiplication by the sum being finite, since is assumed to be locally finite. Using these definitions, it is not hard to show that is a noncommutative algebra, called the incidence algebra of . The identity in this algebra is Appendix: Mo bius Inversion 321 if if The next theorem characterizes those elements of inverses. Theorem A.2.1 An element for all . Proof. A right inverse of that have multiplicative is invertible if and only if must satisfy (A.2.1) In particular, for , we get 1 This shows the necessity and also that 1 must satisfy (A.2.2) has cardinality , that for intervals of all Equation (A.2.2) defines when the interval is, when . We can use (A.2.1) to define cardinalities. Suppose that , and let has been defined for all intervals with cardinality at most have cardinality . Then, by (A.2.1), since , we get But is defined for we can use this to define since . has cardinality at most , and so Similarly, we can define a left inverse using the analogous process. But and so is an inverse for . , defined by if if Definition The function is called the zeta function. Its inverse is called the Mobius function. The next result follows from the appropriate definitions. 322 Field Theory Theorem A.2.2 The Mobius function is uniquely determined by either of the following conditions: 1) and for , 2) and for , Now we come to the main result. Theorem A.2.3 (Mobius Inversion) Let be a locally finite partially ordered set with zero element . If and are functions from to the field , then (A.2.4) If is a locally finite partially ordered set with , then (A.2.5) Proof. Since all sums are finite, we have, for any , The rest of the theorem is proved similarly. The formulas (A.2.4) and (A.2.5) are called Mobius inversion formulas. Example A.2.2 (Subsets) Let be the set of all subsets of a finite set , partially ordered by set inclusion. We will use the notation for subset and for proper subset. The zeta function is Appendix: Mo bius Inversion 323 if otherwise The Mo bius function is computed as follows. From Theorem A.2.2, we have and So, for , we have It begins to appear that the values of alternate between + and and that Asume this is true for and let . Then Now let be "properties" that the elements of a set may or may not possess, that is, . For , let be the number of elements of that have properties for , and no others. Let be the number of elements of that have at least properties , for . Thus, for , where an empty intersection is defined to be , and Then 324 Field Theory Hence, by Mo bius inversion, that is, In particular, if is the empty set, then where Since is the number of elements of that have none of the properties. and since the first term in the previous expression for is , we get For example, if , then This formula is the wellknown Principle of InclusionExclusion, which we now see is just a special case of Mo bius inversion. Classical Mobius Inversion
Consider the partially ordered set of positive natural numbers, ordered by division. That is, is less than or equal to if and only if divides , which we will denote by . The zero element is . In this case, the Mo bius function given by depends only on the ratio / , and is Appendix: Mo bius Inversion 325 if if otherwise Notice that the "otherwise" case can occur if either if 2 for some prime . for distinct primes ( does not divide ) or To verify that this is indeed the Mo bius function, we first observe that . Now let and where the are distinct primes. Then Now, in the present context, the Mo bius inversion formula becomes This is the important classical formula, which often goes by the name Mo bius inversion formula. Multiplicative Version of Mobius Inversion
We now present a multiplicative version of the Mo bius inversion formula. Theorem A.2.4 Let be a locally finite partially ordered set with zero element . If and are functions from to , then 326 Field Theory Proof. Since all products are finite, we have, for any , Example A.2.3 Let Consider the formula , and let be the field of rational functions in . Then, if we let and , Theorem A.2.4 gives References Artin, E., Galois Theory, 2nd ed., Notre Dame Press, 1959. Bescovitch, A.S., On the linear independence of fractional powers of integers, J. London Math. Soc. 15 (1940) 36. Brawley, J. and Schnibben, G., Infinite Algebraic Extensions of Finite Fields, AMS, 1989. Edwards, H., Galois Theory, SpringerVerlag, 1984. Gaal, L., Classical Galois Theory, 4th ed., Chelsea, 1988. Jacobson, N., Basic Algebra II, Freeman, 1989. Lidl, R. and Niederreiter, H., Introduction to Finite Fields and Their Applications, Cambridge University Press, 1986. Richards, I., An application of Galois theory to elementary arithmetic, Advances in Mathematics 13 (1974) 268273. Roman, S., Advanced Linear Algebra, second edition, SpringerVerlag, 2005. Schinzel, A., Abelian binomials, power residues, and exponential congruences, Acta. Arith. 32 (1977) 245274. Index 2tower 44 abelian 3,164,269 abelian series 165 addition 10 affine transformation 285 algebraic 32,43,55 algebraic closure 57 algebraic closure of 55 algebraic integers 56 algebraic numbers 55,67 algebraically closed 56 algebraically dependent on 96 algebraically dependent over 98 algebraically independent 206 algebraically independent of 96 algebraically independent over 98 alternating group 10 antisymmetry 319 antitone 137 ArtinSchreier 266 associates 15 automorphism 11 base 94 bihomomorphism 311 bilinear 17 bilinear form 17,202 binomials 239 biquadratic 91 bottom element 2 Casus Irreducibilis 283 Cauchy's Theorem 9 center 3 chain 1 character 62 characteristic 13 Chinese Remainder Theorem 38 closed 138 Closed fields 149 Closed groups 149 closed under arbitrary composites 45 closure operation 137 closure point 152,153 Closure under finite composites 44 commutative 3 commutative ring 10 commutator subgroup 171,308 commutators 170,308 complete lattice 2 completely closed 142 composite 42 congruent 203 congruent modulo 6 conjugate 154,285 conjugates 33 constructible 69,70,259 Constructible numbers 70 content 24,25 Correspondence Theorem 8 coset 11 cyclic 3,164,269 cyclic subgroup generated by 3 cyclotomic extension of order 241 cyclotomic polynomial 242 De Morgan's Laws 138 Dedekind independence theorem 63 defining polynomial 213 degenerate 202 degree 23,41,140 degree of inseparability 86 degreewise factorization 26 degreewise purely inseparable 87 degreewise reducible 26 degreewise separable 77 dependence relation 93 dependent 94 dependent on 93 330 Field Theory distinguished 44 divides 14 division algorithm 27 dual basis 209 Eisenstein's criterion 37 elementary symmetric polynomials 117,175 embedding 11,58 endomorphism 11 epimorphism 7,11 essentially unique 16 Euclidean domain 17 Euler phi function 6 Euler's Theorem 6 exponent 3,4,273,309 exponent characteristic 239 extension 11,23,58 Extensive 137 factor ring 12 factorization property 16 FeitThompson 270 Fermat's Theorem 6 field 10 field of quotients 15 field primitive element 214 field table 226 finite 3,53 finite extension 43,53,141 finite topology 153 finitely generated 43,46 First Isomorphism Theorem 8,12 five basic operations 281 fixed field 90,143 free from 171 Frobenius automorphism 215 Frobenius map 14 full extension 44 fundamental theorem of algebra 179 Fundamental Theorem of Galois Theory 147,149,156 Galois 148 Galois connection 137 Galois correspondence 144 Galois extension 148 Galois group 143 Galois group of over 143 Galois group of a composite 162 Galois group of a lifting 159,161 Galois group of a polynomial 173 Galois resolvent 129 Galoisstyle group 131 Gauss's lemma 25 generates 3 generic polynomial 117,174 greatest common divisor 15,29 group 2 group homomorphism 7 group primitive element 214 Hilbert's Theorem 90 263 homomorphism 11,62 ideal 11 ideal generated 11 Idempotent 137 incidence algebra 320 independent 94,166 independent of 94 index 4,5,140,141 indexed 140 induced inverse map 8 induced map 8 inseparable 33,73,77 inseparable degree 86 integral domain 10 intermediate field 44 invariant 63 invariant under 63 irreducible 15,24 is closed 143 isomorphic 7 isomorphism 7,11 Isotone 138 join 2 kernel 7 Krull topology 154 Kummer extension 310 Index 331 Lagrange 4 Lagrange's theorem 9 lattice 2 leading coefficient 23 least upper bound 1 left cosets 4 left kernel 311 lifting 44 Lifting Property 44 linearized polynomial 236 linearly disjoint 166 linearly ordered set 320 localization 36 localization maps 24 locally finite 320 lower step 44 Luroth's Theorem 105 matrix of the form 203 maximal 12 maximal element 1 meet 2 metric vector space 202 minimal element 1 minimal polynomial 32 Mobius function 321 Mobius Inversion 322 monic 23 monoid 62 monomial 42 monomorphism 7,11 multiple root 33 multiplication 10 multiplicity 33 natural projection 7 Newton identities 178 Newton's identities 194 Newton's Theorem 118,176 nondegenerate 202 nonsingular 202,312 nontrivial polynomial relationship 99 norm 198 normal 5 normal basis 207,219 normal closure 66 normal extension 64 normal over 64 normal series 135,269 normalizer 5 th roots of unity 119 obtainable by formula 281 obtainable by formula from 281 order 3,170,216,217 orderreversing 137 over 26,58,171 pairing 311 partially ordered set 1,319 perfect 84 perfect closure 89 permutation polynomial 215 pid 16 polynomial 236 polynomial basis 207,209,218 poset 1 prime 12,15 prime subfield 13 primitive 24,25,119,216,240 primitive element 43,46 primitive root modulo 125 principal ideal 11 principal ideal domain 16 properly divides 14 pure 262,265 pure 273 pure 273 pure 273 purely inseparable 85 purely inseparable closure 89 Purely inseparable elements 87 Purely inseparable extensions 87 purely inseparably generated 87 purely transcendental 101 quadratic extension 67,70 quadratic tower 70 quotient group 5 radical exponent 34,73 332 Field Theory radical extension 273 radical series 273 realizable 253 reciprocal polynomial 39 reduced polynomial 187 reducible 24 reducing polynomial 229 reflexivity 319 relatively prime 6,15,31 resolvent equation 122 resolvents 122 right cosets 4 right kernel 311 ring 10 roots of unity over 239 Second Isomorphism Theorem 8 selfreciprocal 39 separable 33,73,77 separable closure 80,89 separable degree 76 separably generated 77 set product 3,4 simple 5,43,46 simple root 33 singular 202 soluble 269 solvable 135,165,269,270 solvable by radicals 274 splits 31,32 splitting field 32,63 standard form 237 Steinitz exchange axiom 94 Steinitz number 220 subfield 11 subgroup 3 sublattice 2 subring 11 Sylow subgroup 9 Sylow's Theorem 9 symmetric 117,176,202 symmetric group 10 tensor product 18 theorem of the primitive element 81 theorem on natural irrationalities 280 Third Isomorphism Theorem 8 top element 2 totally degenerate 202 totally ordered 1 totally ordered set 320 totally singular 202 tower 41 Tower Property 44 trace 198 transcendence basis 100 transcendence degree 100 transcendental 32,43,55,96 transitive 10 transitivity 319 translations 285 transposition 10 type 262,265,273 ufd 16 unique factorization domain 16 unit 10 upper bound 1 upper step 44 Wedderburn's Theorem 251 Wilson's theorem 223 zero divisor 10 zeta function 321 Zorn's lemma 2 ...
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